Differentiation Part4

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Parametric Functions

Remember that parametric functions will be in the form


Notice that normally, when we have {\color{Red}y}=f\left({\color{Blue}x}\right), we can find \frac{d{\color{Red}y}}{d{\color{Blue}x}}

Imagine we have {\color{Red}\heartsuit}=f\left({\color{Blue}\bigcirc}\right), we can find \frac{d{\color{Red}\heartsuit}}{d{\color{Blue}\bigcirc}}

Thus, since we have x=f\left(t\right), y=f\left(t \right), we can find

But how would be find \frac{dy}{dx}?

Remember that chain rule give us \frac{dy}{dx}=\frac{dy}{d{\color{Red}\heartsuit}}\times\frac{d{\color{Red}\heartsuit}}{dx}

Since we already have \frac{dx}{dt} and \frac{dy}{dt}, thus we will use

However notice that we have can find \frac{dx}{dt} but we need \frac{dt}{dx}

Thus we can use the law either as

However, if we have fraction for \frac{dx}{dt}, it's better we just multiply the reciprocal (rather than divide by a fraction)

Note:

Quick Practice:


Beyond that, there isn't really anything new you need to learn here. However, the following examples and exercises WILL serve as a very good practice of identifying what kind of rules we will use to get the individual derivatives , since there are all mixed up here compared to previous exercises.

Examples

Exercise 7

Find \frac{dy}{dx} for the following

a) x=4t^{2}-1,\;y=t^{3}-2t \frac{dy}{dx}=\frac{3t^{2}-2}{8t}

b) x=3\left(t^{2}-3\right)^{6},\;y=1-t^{3} \frac{dy}{dx}=-\frac{t}{12\left(t^{2}-3\right)^{5}}

c) x=t-\frac{1}{t},\;y=t+\frac{1}{t} \frac{dy}{dx}=\frac{t^{2}-1}{t^{2}+1}

d) x=\sqrt{t},\;y=\frac{3}{t^{3}-1} \frac{dy}{dx}=-\frac{18t^{\frac{3}{2}}}{\left(t^{3}-1\right)^{2}}

e) x=3\cos t,\;y=3\sin t \frac{dy}{dx}=-\frac{\cos t}{\sin t}

f) x=\sin^{4}2\theta ,\;y=\cos^{4}2\theta = \frac{dy}{dx}=-\frac{\cos^{2}2\theta }{\sin^{2}2\theta }

g) x=\ln\left(t^{2}+1\right),\;y=\ln\left(e^{t}\sin 3t\right) \frac{dy}{dx}=\frac{\left(t^{2}+1\right)\left(\sin 3t+3\cos 3t\right)}{2t\sin 3t}

h) x=\frac{e^{t}-e^{-t}}{e^{2t}},\;y=\ln\sqrt{\frac{5}{t^{5}}} \frac{dy}{dx}=-\frac{5}{2t\left(-e^{-t}+3e^{-3t}\right)}

i) x=t^{2}e^{2t},\;y=\frac{e^{2t}}{t^{2}} \frac{dy}{dx}=\frac{\left(t-1\right)}{t^{4}\left(t+1\right)}

j) x=\sqrt{e^{t}}+\frac{1}{\sqrt{e^{t}}},\;y=\sqrt{e^{t}+4} \frac{dy}{dx}=\frac{e^{\frac{3}{2}t}}{\sqrt{e^{t}+4}\left(e^{t}-1\right)}

Implicit Functions

Up to now, we have dealt with functions which is given in the form of y=f(x)\,, which are explicit functions.

Implicit functions are simply functions/relations NOT written explicitly in the form of y=f(x)\,.

Note :

Derivative of f(y)


\frac{d\left[f\left({\color{Blue}y}\right)\right]}{d{\color{Red}x}} =\frac{d\left[f\left({\color{Blue}y}\right)\right]}{?}\times\frac{?}{d{\color{Red}x}}

In other words

Once again

Examples

Products/Quotients

As before, basically, just remember that products/quotients takes precedence (as in we apply the products/quotients formulas first before anything else). In fact, there is nothing new to learn here, just a practice of sticking to the formulas even when things get very complicated (and as always, learning how to check to make sure we didn't miss out anything)

Let's try

Examples

Differentiate the following w.r.t. x\,. Hence. find \frac{dy}{dx}

Exercise 8

Differentiate the following with respect to x\,. Hence, find \frac{dy}{dx}.

a) x^{2}-y^{2}=5\,

b) \sqrt{x}+\sqrt{y}=2

c) xy+\sin x=0\,

d) x^{2}y^{3}=e^{x}\,

f) \sin y+\sin x=1\,

g) \ln \left(y+4\right)=4x

h) e^{x+y}=\sin 2x\,

i) \sin\left(xy\right)=x\,

j) \sin^{2}y=e^{x^{2}}

Derivatives using logarithm

Prove of \frac{d}{dx}{a}^{x} ={a}^{x}\ln a \,

Note: Alternatively, we can prove this formula using the identity e^{\ln x}=x\, and \frac{d}{dx}\left(e^{x}\right)=e^{x}


Notice that the method works because logarihms allows us to bring down the index to make the differentiating much easier. Additionally, we can also use it to change times/divide to plus/minus.

Thus, we can use this method to as alternative method to solve questions with difficult index, such as \frac{3x-2}{\sqrt[3]{x+1}} (actually, the differentiation itself isn't that difficult even if we use the normal method, but the simplifying after that can be tricky)

Note:

Example

Using logarithm, or otherwise, differentiate y=\frac{3x-2}{\sqrt[3]{x+1}} w.r.t. x\,

Alternatively,

Exercise 9

Using logarithm, or otherwise, find \frac{dy}{dx} for the following.

a)y=\sqrt[4]{3x-1}\left(2x-3\right)

OR


b)y=\frac{\left(x-1\right)^{4}}{\sqrt[3]{x+1}}

OR

Higher Derivatives

If we differentiate \frac{dy}{dx} w.r.t. x\,, what do we get?

If we keep on differentiation, the notation will be as follows

Example

Proving Differential Equations

These are a special type of question whereby we are given a function, and instead of finding the actual formula for its derivative(s), we are asked to prove that the function satisfies a given differential equation.

Example : If y=e^{-x}\left(ax+b\right), where a, b\, are constants, prove that \frac{d^{2}y}{dx^2}+2\frac{dy}{dx}+y=0

General Guidelines :

Each question, is of course, different. But always keep the following two tips in mind when attempting this kind of question

Examples

Exercise 10

1)If y=e^{3x}\left(9x+1\right), prove that \frac{d^{2}y}{dx^{2}}-6\frac{dx}{dy}+9y=0

2)If y=\frac{\sin x}{x}, prove that x\left(\frac{d^{2}y}{dx^{2}}\right)+2\frac{dx}{dy}+xy=0

3)If y=\frac{1}{1+x^{2}}, prove that \left(1+x^{2}\right)\frac{d^{2}y}{dx^2}+4x\frac{dy}{dx}+2y=0

4)If y=\ln\left(\sin x+\cos x\right), prove that \frac{d^{2}y}{dx^{2}}+\left(\frac{dx}{dy}\right)^{2}+1=0

5)If ye^{mx}=a\cos px\,, where a, m, p\, are constants, prove that \frac{d^{2}y}{dx^{2}}+2m\frac{dx}{dy}+\left(m^{2}+p^{2}\right)y=0

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