Differentiation Part1

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Contents

Notes

Learning Objectives (Syllabus)

  • use the notations f'(x), f''(x), \frac{dy}{dx},\frac{{d}^{2}y}{d{x}^{2}}
  • use the derivative of {x}^{n}\, (for any rational number n\,),{e}^{x},\ln x, \sin x, \cos x, \tan x \,
  • carry out differentiation of kf(x),f(x)\pm g(x),f(x)g(x),\frac{f(x)}{g(x)},\left(f\circ g\right)(x)
  • find the first derivative of an implicit function
  • find the first derivative of a function defined parametrically

Prior Knowledge

  • Strong basics in index is a MUST in this topic
  • Strong in algebraic manipulations (factoring, rearranging, fractions, etc) is also a MUST for simplifying

Notation

When we differentiate y=f(x)\, with respect to (w.r.t.) x\, we will obtain the derivative which can be written in any of the following form

  • \frac{dy}{dx}
  • f'\left(x\right)
  • \frac{d}{dx}\left[f\left(x\right)\right]

Basic Formulas

Constants

A) \frac{d}{dx}c =0\, (c\, is a constant)

  • Remember : Derivative of ANY constant is zero
  • Note :
    • Which of the following are constants? 1,-50,\frac{1}{2}, \ln 2, \pi, e, \frac{\pi}{100},{e}^{2}+2. ALL
    • Any other "alphabets" may/may not be constants (it will be stated in the question)

X power n

B) \frac{d}{dx}{x}^{n} =n{x}^{n-1}\,

  • Remember : "Bring" down the index, minus one to the index
  • Note :
    • Terms like \frac{1}{x},\sqrt{x},\sqrt[3]{x},\frac{1}{\sqrt{x}} can be changed to the {x}^{n}\, form before differentiating

Examples

  • \frac{d}{dx}\left({x}^{6}\right) =6{x}^{5}\, \frac{d}{dx}\left({x}^{{\color{Red}6}}\right)={\color{Red}6}{x}^{{\color{Blue}5}}
  • \frac{d}{dx}\left({x}^{-10}\right) =-10{x}^{-11}\, Be careful, -10-1\, is -11\, not -9\,
  • \frac{d}{dx}\left(x\right) =1\, \frac{d}{dx}\left(x\right)=\frac{d}{dx}\left({x}^{{\color{Red}1}}\right)={\color{Red}1}{x}^{{\color{Blue}0}}=1 Of course, the working is not needed next time and we just memorize the formula for this special case
  • \frac{d}{dx}\left(\frac{1}{x}\right) =-\frac{1}{{x}^{2}} \frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}\left({x}^{{\color{Red}-1}}\right)={\color{Red}-}{x}^{{\color{Blue}-2}}=-\frac{1}{{x}^{2}} Note that in the step \frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}\left({x}^{-1}\right) we have only changed the form, and have not differentiated it yet.
  • \frac{d}{dx}\left(\sqrt{x}\right) =\frac{1}{2\sqrt{x}} \frac{d}{dx}\left(\sqrt{x}\right)=\frac{d}{dx}\left({x}^{{\color{Red}\frac{1}{2}}}\right)={\color{Red}\frac{1}{2}}{x}^{{\color{Blue}-\frac{1}{2}}}=\frac{1}{2\sqrt{x}}
  • \frac{d}{dx}\left(\frac{1}{{x}^{4}}\right) =-\frac{4}{{x}^{5}} If you can sort of work out the working in your mind already (though you should spend an extra second to check it), the working is not needed. \frac{d}{dx}\left(\frac{1}{{x}^{4}}\right)=\frac{d}{dx}\left({x}^{{\color{Red}-4}}\right)={\color{Red}-4}{x}^{{\color{Blue}-5}}=-\frac{4}{{x}^{5}}
  • \frac{d}{dx}\left(\frac{1}{\sqrt[3]{{x}^{2}}}\right) =-\frac{2}{3{x}^{\frac{5}{3}}} \frac{d}{dx}\left(\frac{1}{\sqrt[3]{{x}^{2}}}\right)=\frac{d}{dx}\left({x}^{{\color{Red}-\frac{2}{3}}}\right)={\color{Red}-\frac{2}{3}}{x}^{{\color{Blue}-\frac{5}{3}}}=-\frac{2}{3{x}^{\frac{5}{3}}}

Useful Formulas

These should be memorized

  • \frac{d}{dx}\left(x \right)=1 Thus, \frac{d}{dx}\left(cx \right)=c (if c\, is constant) - refer below
  • \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{{x}^{2}}
  • \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt{x}}

Plus/Minus/Multiply with Constant

C) \frac{d}{dx}\left[f\left(x\right)\pm g\left(x\right)\right] =\frac{d}{dx}\left[f\left(x\right)\right]\pm\frac{d}{dx}\left[g\left(x\right)\right]

  • Remember : We differentiate the terms one by one when it is plus/minus

D) \frac{d}{dx}c\left[f\left(x\right)\right] =c\frac{d}{dx}\left[f\left(x\right)\right](c\, is a constant)

  • Remember : When we have a constant times a function, we just leave the constant outside.
  • Note : Note the difference between
    • \frac{d}{dx}c =0\,
    • \frac{d}{dx}c\left[f\left(x\right)\right] =c\frac{d}{dx}\left[f\left(x\right)\right]
    • differentiate constant =0\,
    • differentiate constant times function \rightarrowleave the constant outside
  • Note : C) and D) are general formula can applies for all types of functions

Example

y=3{x}^{2}+\left(\pi+1\right)x-\frac{2}{x}+{\pi}^{2}-4{a}^{3}+1 where a\, is a constant.

  • \frac{dy}{dx}=6x+\left(\pi+1\right)+\frac{2}{{x}^{2}}
    • Analyze : There are multiple terms, but all of it plus/minus, thus we differentiate one by one
    • \frac{d}{dx}\left(3{x}^{2}\right)=3\frac{d}{dx}\left({x}^{2}\right)=3\left(2x\right)=6x. Of course, we don't actually write this working out, instead we just keep in mind that the 3\, is just left as it is (that is, it need not be differentiated)
    • For \left(\pi+1\right)x, it is a constant times a function (x\,). Thus this will just become \left(\pi+1\right)
    • Differentiate \frac{1}{x} is -\frac{1}{{x}^{2}}, thus it becomes {\color{Red}+}\frac{2}{{x}^{2}}
    • {\pi}^{2}\, is a constant. 4{a}^{3}\, is a constant. Thus the last 3 terms will just give us 0\,
    • Thus, \frac{dy}{dx}=6x+\left(\pi+1\right)+\frac{2}{{x}^{2}}

Writing the Working

When writing the working out, always remember the function and its derivative are two DIFFERENT things. Thus, we must write clearly (and be clear about it ourselves) when we actually differentiate (as oppose to rearranging/simplifying) the given function.

  • Example 1 : \begin{align} y &=6x^{2} \\ & = 12x \end{align}
    • Correct? WRONG 6x^{2}\, is definitely NOT EQUAL to 12x\,.
    • It MUST be written as \begin{align}y &=6x^{2} \\ \frac{dy}{dx}& = 12x \end{align}

  • Example 2 : \begin{align} y &=\sqrt{x}\\ &= {x}^{\frac{1}{2}}\\ &=-\frac{1}{2}{x}^{-\frac{1}{2}}\\ &=-\frac{1}{2\sqrt{x}} \end{align}
    • How should we correct it? \begin{align} y &=\sqrt{x}\\ &= {x}^{\frac{1}{2}}\\ \frac{dy}{dx}&=-\frac{1}{2}{x}^{-\frac{1}{2}}\\ &=-\frac{1}{2\sqrt{x}} \end{align}
    • Notice that \begin{align} y &={\color{Red}\sqrt{x}}\\ &= {\color{Red}{x}^{\frac{1}{2}}}\\ \frac{dy}{dx}&={\color{Blue}-\frac{1}{2}{x}^{-\frac{1}{2}}}\\ &={\color{Blue}-\frac{1}{2\sqrt{x}}} \end{align} The red parts was rearranging it before differentiating. The blue parts is simplifying after differentiating. There was only a single step where we actually differentiated it.

  • Example 3 : \begin{align} y &={x}^{2}+\sqrt{x}\\ \frac{dy}{dx}&= 2x+{x}^{\frac{1}{2}}\\ &=2x-\frac{1}{2}{x}^{-\frac{1}{2}}\\ &=2x-\frac{1}{2\sqrt{x}} \end{align}
    • WRONG. Simple terms can be easily differentiated, some might need some rearranging, but we must differentiate all terms TOGETHER in a single step of working.
    • \begin{align} y&={\color{Red}{x}^{2}+\sqrt{x}}\\ &= {\color{Red}{x}^{2}+{x}^{\frac{1}{2}}}\\ \frac{dy}{dx}&={\color{Blue}2x-\frac{1}{2}{x}^{-\frac{1}{2}}}\\ &={\color{Blue}2x-\frac{1}{2\sqrt{x}}} \end{align}

e^x, ln x, sin x, cos x, tan x

Note : We don't really need to know the why's of each formula below, just need to memorize it.

E) \frac{d}{dx}{e}^{x} ={e}^{x}\,

  • Remember : differentiate {e}^{x}\, we get back {e}^{x}\, itself.

F) \frac{d}{dx}{a}^{x} ={a}^{x}\ln a \,

  • Remember : a\, here refers to any other positive constants. Example : \frac{d}{dx}\left({2}^{x}\right) ={2}^{x}\ln 2 \,

G) \frac{d}{dx}\ln x =\frac{1}{x}

  • Remember : differentiate \ln x\, we get \frac{1}{x}

H) \frac{d}{dx}\sin x =\cos x \,

  • Remember : differentiate \sin x\, we get \cos x\,

I) \frac{d}{dx}\cos x =-\sin x \,

  • Remember : differentiate \cos x\, we get {\color{Red}-}\sin x\,

J) \frac{d}{dx}\tan x ={\sec}^{2}x \,

  • Remember : differentiate \tan x\, we get ={\sec}^{{\color{Red}2}}x

Examples

Find f'\left(x\right) for the following

  • f\left(x\right)={e}^{x}+\ln x
    • f'\left(x\right)={e}^{x}+\frac{1}{x}
  • f\left(x\right)=2\sin x-\cos x + \frac{1}{3}\tan x
    • f'\left(x\right)=2\cos x+\sin x + \frac{1}{3}{\sec}^{2}x
  • f\left(x\right)=2\left(3{x}^{4}-2\sin x +5 \right)
    • f'\left(x\right)=2\left(12{x}^{3}-2\cos x \right)
    • There is not a need to multiply the 2\, inside before differentiating. Remember that \frac{d}{dx}c\left[f\left(x\right)\right] =c\frac{d}{dx}\left[f\left(x\right)\right] and this applies for ANY f\left(x\right). Thus, for the question above, just leave the 2\, outside and differentiate whatever is inside the bracket as usual.

Exercise 1

Differentiate the following w.r.t. x\,

a) f\left(x\right)={x}^{7}+5{x}^{2}-3{x}^{-2}+3{x}^{\frac{1}{2}}

  • f'\left(x\right)=7{x}^{6}+10x+6{x}^{-3}+\frac{3}{2}{x}^{-\frac{1}{2}}

b) f\left(x\right)=2{x}^{5}-4x+\frac{3}{x}+5\sqrt{x}-\frac{2}{\sqrt{x}}

  • f'\left(x\right)=10{x}^{4}-4-\frac{3}{{x}^{2}}+\frac{5}{2\sqrt{x}}+\frac{1}{{x}^{\frac{3}{2}}}

c) f\left(x\right)=2\ln x-{e}^{x}+{2}^{x}-{3}^{x}

  • f'\left(x\right)=\frac{2}{x}-{e}^{x}+{2}^{x}\ln 2-{3}^{x}\ln 3

d) f\left(x\right)=4\tan x+7\cos x-3\sin x

  • f'\left(x\right)=4{\sec}^{2}x- 7\sin x-3\cos x

e) f\left(x\right)=\frac{2}{{x}^{4}}-\frac{\ln x}{4}+\frac{2\tan x}{3}

  • f'\left(x\right)=-\frac{8}{{x}^{5}}-\frac{1}{4x}+\frac{2}{3}{\sec}^{2}x

Products and Quotients

K) if y=uv\,, where u,v\, are functions of x\,

  • \frac{dy}{dx} =u\frac{dv}{dx}+v\frac{du}{dx}
    • Remember : When we differentiate something times something, we copy the first thing multiply differentiate the second thing, and then plus copy the second thing, differentiate the first thing.

L) if y=\frac{u}{v}, where u,v\, are functions of x\,

  • \frac{dy}{dx} =\cfrac{v\cfrac{du}{dx}-u\cfrac{dv}{dx}}{{v}^{2}}
    • Remember : When we differentiate something over something, we copy the bottom thing multiply differentiate the top thing, and then MINUS copy the top thing, differentiate the bottom thing, the whole thing OVER bottom thing square.
  • Important Note: u,v\, are functions of x\,, if constants, the formulas are NOT needed (just use formula D)
    • Example : \left(x+2\right)\left(x-2\right), x{e}^{x}, {x}^{2}{\sin x}, \frac{\tan x}{x} All these we would require the above formulas (though the first one can be done alternatively).
    • Example : 3{e}^{x},\pi \sin x, \frac{\tan x}{3} All these DOES NOT require the above formulas.

Differentiating Complicated Functions

When u,v\, are complicated functions, we can differentiate it separately away from the main workings.

Say, for example, we have {e}^{\sin x}\cos \left(ax+b\right), we can see it's something times something {\color{Red}{e}^{\sin x}}{\color{Blue}\cos \left(ax+b\right)}. Differentiating those parts aren't really that difficult once we get the hang of it, but it's very easy to make careless mistakes if we do it too quickly. My suggestion is to do some side working.

  • One way we could do is \frac{dy}{dx} ={e}^{\sin x}\frac{d}{dx}\left[ \cos \left(ax+b\right)\right]+\cos \left(ax+b\right)\frac{d}{dx}\left({e}^{\sin x}\right) but I find this to be more confusing to students most of the time
  • Instead, we do it as side working
    • First we write \frac{dy}{dx}={e}^{\sin x}
    • Next we need to differentiate \cos \left(ax+b\right) and put it after \frac{dy}{dx}={e}^{\sin x}, so we can do it as side working and then put whatever we get into it, and so on.
    • \begin{array}{ll} y={e}^{\sin x}\qquad\qquad\qquad\qquad\qquad\quad & \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit} \end{array}
    • \begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)\qquad\quad\qquad\qquad\qquad\;& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit} \end{array}
    • \begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)+\left[\cos\left(ax+b\right)\right]\qquad\;& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit} \end{array}
    • \begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)+\left[\cos\left(ax+b\right)\right]\qquad\;& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit}\quad \cfrac{d}{dx}{e}^{\sin x}={\color{Blue}\bigcirc} \end{array}
    • \begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)+\left[\cos\left(ax+b\right)\right]{\color{Blue}\bigcirc}\quad& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit}\quad \cfrac{d}{dx}{e}^{\sin x}={\color{Blue}\bigcirc} \end{array}
    • This method helps us concentrate on one thing at one time and also easy for us to check quickly for careless mistakes

Examples

Differentiate the following w.r.t x\,

  • y=x\sin x\, \frac{dy}{dx}=x\cos x+\sin x
    • Analyze : Something times something. No way to simplify.
      • Copy first thing : \frac{dy}{dx}={\color{Red}x}
      • Multiply differentiate second thing: \frac{dy}{dx}={\color{Red}x}{\color{Blue}\cos x}
      • Plus copy second thing: \frac{dy}{dx}={\color{Red}x}{\color{Blue}\cos x}+{\color{Blue}\sin x}
      • Multiply differentiate first thing: \frac{dy}{dx}={\color{Red}x}{\color{Blue}\cos x}+\left({\color{Blue}\sin x}\right){\color{Red}1}=x\cos x+\sin x
        • Note : Proper usage of brackets are important. It is \left(\sin x \right)1, and NOT \sin x\left(1 \right), since the 1\, multiplies the \sin x \,, and not the x\,. When in doubt, just put brackets for both, i.e. \left(\sin x \right)\left(1 \right)

  • y=\left({x}^{2}+1\right)\ln x \frac{dy}{dx}=\frac{{x}^{2}+1}{x}+2x\ln x
    • Analyze : Something times something. {\color{Red}\left({x}^{2}+1\right)}{\color{Blue}\ln x}
      • Put brackets where appropriate (or at both if in doubt)
        • \frac{dy}{dx}=\left({x}^{2}+1\right)\left(\frac{1}{x}\right)+\ln x\left(2x \right) Correct? No. Since \ln x\left(2x \right) would be misunderstood as \ln 2{x}^{2}\,
        • \frac{dy}{dx}=\left({x}^{2}+1\right)\left(\frac{1}{x}\right)+\left(\ln x \right)2x= \frac{{x}^{2}+1}{x}+2x\ln x Note : Brackets are not needed when writing 2x\ln x\, since there is no danger of misinterpration

  • y=\frac{{e}^{x}}{x} \frac{dy}{dx}=\frac{{e}^{x}\left(x-1\right)}{{x}^{2}}
    • Analyze : Something over something. Can't simplify.
      • \frac{dy}{dx}=\frac{\qquad}{\qquad}
        • Copy bottom differentiate top \frac{dy}{dx}=\frac{{\color{Red}x}{\color{Blue}{e}^{x}}}{\qquad}
        • Minus copy top differentiate bottom \frac{dy}{dx}=\frac{{\color{Red}x}{\color{Blue}{e}^{x}}-{\color{Blue}{e}^{x}}}{\qquad} (don't need to write the 1\,)
        • Divide by bottom square \frac{dy}{dx}=\frac{{\color{Red}x}{\color{Blue}{e}^{x}}-{\color{Blue}{e}^{x}}}{{{\color{Red}x}}^{2}}
        • Simplify if possible. Factor. \frac{dy}{dx}=\frac{{e}^{x}\left(x-1\right)}{{x}^{2}}
  • y=\frac{2x+1}{\ln x} \frac{dy}{dx}=\frac{2x\ln x+2x+1}{x{\left(\ln x\right)}^{2}}
    • Analyze : Something over something. No point simplifying (breaking the fractions apart).
      • Differentiate as usual : \frac{dy}{dx}=\frac{\left(\ln x\right)2+\cfrac{\left(2x+1\right)}{x}}{{\left(\ln x\right)}^{2}}
      • How should we simplify it? We change it to a single fraction.
        • Compare with \frac{a+{\frac{b}{c}}}{d}
          • We first deal with the top part to make it a single fraction \frac{\left(\cfrac{ac+b}{c}\right)}{d}, thus it will become \frac{ac+b}{cd}
          • Thus, next time we just do it directly, \frac{a+{\frac{b}{{\color{Red}c}}}}{d} =\frac{a{\color{Red}c}+b}{{\color{Red}c}d}
      • Using the same method, \frac{dy}{dx}=\frac{2{\color{Red}x}\ln x+2x+1}{{\color{Red}x}{\left(\ln x\right)}^{2}}

  • y=\frac{{e}^{x}}{\sqrt{x}} \frac{dy}{dx}=\frac{{e}^{x}\left(2x-1\right)}{2{x}^{\frac{3}{2}}}
    • Analyze : Something over something.
      • Notice that we are going to differentiate \sqrt{x}, it's the best that we use the formula \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt{x}} DIRECTLY. Even if we have forgotten the formula, work it out at the side.
      • Thus, differentiating, we get \frac{dy}{dx}=\frac{\sqrt{x}\left({e}^{x}\right)-{e}^{x}\left(\cfrac{1}{2\sqrt{x}}\right)}{{\left(\sqrt{x}\right)}^{2}}
      • If we had chosen to do the derivative of \sqrt{x} right in the middle of everything, we will get \frac{dy}{dx}=\frac{{x}^{\frac{1}{2}}\left({e}^{x}\right)-{e}^{x}\left(\frac{1}{2}{x}^{-\frac{1}{2}}\right)}{{\left({x}^{\frac{1}{2}}\right)}^{2}} which is not just more messy, but makes it more difficult to see how we are going to simplify.
      • The bottom part {\left(\sqrt{x}\right)}^{2}=x, and making the whole thing into a single fraction, we will get \frac{dy}{dx}=\frac{2\sqrt{x}\sqrt{x}\left({e}^{x}\right)-{e}^{x}}{2\sqrt{x}\left(x\right)}, but since we know that \sqrt{x}\sqrt{x}=x, we might as well just write it directly
      • \frac{dy}{dx}=\frac{2x\left({e}^{x}\right)-{e}^{x}}{2\sqrt{x}\left(x\right)}
      • Factorizing and combining the bottom parts, \frac{dy}{dx}=\frac{{e}^{x}\left(2x-1\right)}{2{x}^{\frac{3}{2}}}

Exercise 2

Differentiate the following w.r.t. x\,

a) y=\left(3x+1\right)\ln x

  • \frac{dy}{dx}=\frac{\left(3x+1\right)}{x}+3\ln x

b) y={x}^{2}\cos x\,

  • \frac{dy}{dx}=x\left(2\cos x-x\sin x \right)
    • \begin{align} y &={x}^{2}\cos x \\ \frac{dy}{dx} & =-{x}^{2}\sin x+2x\cos x\\ & =x\left(2\cos x-x\sin x \right) \end{align}

c) y ={e}^{x}\sin x\,

  • \frac{dy}{dx}={e}^{x}\left(\cos x+\sin x\right)
    • \begin{align} y &={e}^{x}\sin x \\ \frac{dy}{dx} & ={e}^{x}\cos x +\left( \sin x\right){e}^{x} \\ &= {e}^{x}\left(\cos x+\sin x\right) \end{align}

d) y =x{e}^{x}\cos x \,

  • \frac{dy}{dx}={e}^{x}\left(x\cos x-x\sin x+\cos x \right)
    • \begin{align} y &=x{e}^{x}\cos x \\ &=\left(x{e}^{x}\right)\cos x \\ \frac{dy}{dx} & =-\left(x{e}^{x}\right)\sin x +\left(\cos x\right)\left(x{e}^{x}+{e}^{x} \right) \\ & ={e}^{x}\left(x\cos x-x\sin x+\cos x \right) \end{align} \begin{align} & \mbox{Let } y= x{e}^{x} \\ & \frac{dy}{dx} = xe^{x}+e^{x} \end{align}

e) y =\left({e}^{x}-3\right)\left(\sin x+2x \right)

  • \frac{dy}{dx} =\left({e}^{x}-3\right)\left(\cos x+2\right)+{e}^{x}\left(\sin x+2x \right)

f) y ={e}^{x}\left(2\cos x-3{x}^{2} \right)

  • \frac{dy}{dx}={e}^{x}\left(-2\sin x+2\cos x-3{x}^{2}-6x\right)
    • \begin{align} y &={e}^{x}\left(2\cos x-3{x}^{2} \right) \\ \frac{dy}{dx} & ={e}^{x}\left(-2\sin x-6x\right)+\left(2\cos x-3{x}^{2} \right){e}^{x}\\ &={e}^{x}\left(-2\sin x+2\cos x-3{x}^{2}-6x\right) \end{align}

g) y =\frac{x+1}{x+2}

  • \frac{dy}{dx}=\frac{1}{{\left(x+2\right)}^{2}}
    • \begin{align} y &=\frac{x+1}{x+2} \\ \frac{dy}{dx} & = \frac{\left(x+2\right)-\left(x+1\right)}{{\left(x+2\right)}^{2}}\\ &= \frac{1}{{\left(x+2\right)}^{2}} \end{align}

h) y =\frac{6{x}^{3}}{{x}^{2}-1}

  • \frac{dy}{dx} = \frac{6{x}^{2}\left({x}^{2}-3\right)}{{\left({x}^{2}-1\right)}^{2}}
    • \begin{align} y &=\frac{6{x}^{3}}{{x}^{2}-1} \\ \frac{dy}{dx} & = \frac{\left({x}^{2}-1\right)\cdot 18{x}^{2} -6{x}^{3}\cdot 2x}{{\left({x}^{2}-1\right)}^{2}}\\ &= \frac{6{x}^{2}\left(3{x}^{2}-3-2{x}^{2}\right)}{{\left({x}^{2}-1\right)}^{2}}\\ &= \frac{6{x}^{2}\left({x}^{2}-3\right)}{{\left({x}^{2}-1\right)}^{2}} \end{align}

i) y =\frac{1-x}{\sqrt{x+1}}

  • \frac{dy}{dx}=\frac{-x-3}{2{\left(x+1\right)}^{\frac{3}{2}}}
    • \begin{align} y &=\frac{1-x}{\sqrt{x+1}} \\ \frac{dy}{dx} & = \frac{\sqrt{x+1}\left(-1\right)-\left(1-x\right)\cfrac{1}{2\sqrt{x+1}}}{x+1}\\ & = \frac{-2\left(x+1\right)-\left(1-x\right)}{2\sqrt{x+1}\left(x+1\right)}\\ & = \frac{-x-3}{2{\left(x+1\right)}^{\frac{3}{2}}}\\ \end{align}

j) y =\frac{\ln x}{x}

  • \frac{dy}{dx}=\frac{1-\ln x}{{x}^{2}}
    • \begin{align} y &=\frac{\ln x}{x} \\ \frac{dy}{dx} & = \frac{x\left(\cfrac{1}{x}\right)-\ln x}{{x}^{2}}\\ & = \frac{1-\ln x}{{x}^{2}}\\ \end{align}

k) y=\frac{{e}^{x}}{{x}^{3}}

  • \frac{dy}{dx} =\frac{{e}^{x}\left(x-3 \right)}{{x}^{4}}
    • \begin{align} y &=\frac{{e}^{x}}{{x}^{3}} \\ \frac{dy}{dx} & = \frac{{x}^{3}{e}^{x}-{e}^{x}\cdot 3{x}^{2}}{{x}^{6}}\\ & = \frac{{x}^{2}{e}^{x}\left(x-3 \right)}{{x}^{6}}\\ & = \frac{{e}^{x}\left(x-3 \right)}{{x}^{4}}\\ \end{align}

l) y=\frac{x+1}{\sin x}

  • \frac{dy}{dx} =\frac{\sin x-\left(x+1\right)\cos x}{{\left(\sin x\right)}^{2}}

m) y=\frac{\sin x}{\sqrt{x}}

  • \frac{dy}{dx} =\frac{2x\cos x-\sin x}{2{x}^{\frac{3}{2}}}
    • \begin{align} y &=\frac{\sin x}{\sqrt{x}} \\ \frac{dy}{dx} &=\frac{\sqrt{x}\cos x-\left(\sin x \right)\left(\cfrac{1}{2\sqrt{x}} \right)}{x}\\ & = \frac{2x\cos x-\sin x}{2{x}^{\frac{3}{2}}} \end{align}

n) y=\frac{x}{2\ln x}

  • \frac{dy}{dx} =\frac{\ln x -1}{2{\left(\ln x \right)}^{2}}
    • \begin{align} y &=\frac{x}{2\ln x}\\ \frac{dy}{dx} &=\frac{1}{2}\left[\frac{\ln x-x\left(\cfrac{1}{x}\right)}{{\left(\ln x \right)}^{2}}\right]\\ &=\frac{\ln x -1}{2{\left(\ln x \right)}^{2}} \end{align}
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