HARD QUESTIONS INVOLVING DIFFERENTIATION

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Differentiation is the process of finding the derivative of a function.

In today post, we are going to be trying some questions on differentiation.

Forthwith, let's get started.

Question 1

Differentiate $y=\frac{\ln(x)}{x^3}$

Answer

This is a quotient. Hence, we use quotient rule which is exemplified by below:

$$\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{(g)^2}$$

Here, $f=\ln (x)$, $f'=\frac{1}{x}$, $g=x^3$, $g'=3x^2$.

$\frac{dy}{dx}=\frac{(\frac{1}{x}\times x^3)-(3x^2\times \ln(x))}{(x^3)^2}$

$\frac{dy}{dx}=\frac{\frac{x^3}{x}-3x^2 \ln(x)}{x^6}$

$\frac{dy}{dx}=\frac{x^2-3x^2 \ln(x)}{x^6}$

$\frac{dy}{dx}=\frac{x^2(1- 3\ln(x))}{x^6}$

$\frac{dy}{dx}=\frac{1- 3\ln(x)}{x^4}$

Question 2

Differentiate the function $h(x)=2x-\frac{3}{x^2}+4\sqrt{x}+2$

Answer

We can rewrite the above expression as

$$h(x)=2x-3x^{-2}+4x^{\frac{1}{2}}+2$$

Using power rule

$h'(x)=2x^{1-1}+6x^{-2-1}+\frac{4}{2}x^{\left(\frac{1}{2}-1\right)}+0$

$h'(x)=2+6x^{-3}+2x^{\left(\frac{-1}{2}\right)}$

If you would like to express it in fraction form,

$h'(x)=2+\frac{6}{x^3}+\frac{2}{\sqrt{x}}$

Question 3

Find the derivative of the function $h(x)=x^2\cot 2x$

Answer

$x^2\cot 2x$ is the product of $x^2$ and $\cot 2x$. Therefore, we use the product rule which is shown below

$$(fg)'=f'g+g'f$$

$f=x^2, f'=2x, g=\cot 2x, g'=-\csc^2 2x$

$h'(x)=2x(\cot 2x)+(-\csc^2 2x)(x^2)$

$h'(x)=2x\cot 2x-x^2\csc^2 2x$

$h'(x)=x(2\cot 2x-x\csc^2 2x)$

Question 4

Find the derivative of $y=\frac{\ln(x)}{1+\ln(x)}$

Answer

Using the quotient rule.

Here $f=\ln (x), f=\frac{1}{x}, g=1-\ln (x), g'=\frac{1}{x}$

$\frac{dy}{dx}=\frac{\frac{1}{x}(1+\ln(x))+\frac{1}{x}(\ln (x))}{(1+\ln(x))^2}$

$\frac{dy}{dx}=\frac{\frac{1+in(x)}{x}-\frac{\ln(x)}{x}}{(1+\ln(x))^2}$

$\frac{dy}{dx}=\frac{\frac{1+\ln(x)-\ln(x)}{x}}{(1+\ln(x))^2}$

$\frac{dy}{dx}=\frac{1}{x}\times\frac{1}{(1+\ln(x))^2}$

$\frac{dy}{dx}=\frac{1}{x(1+\ln(x))^2}$

Question 5

Differentiate $\ln(\frac{1}{x^2+9})$

Answer

First, we simplify the natural log

$in(\frac{1}{x^2+9})=\ln(x^2+9)^{-1}$

By the property of logarithm

$\ln(x^2+9)^{-1}

$-1\ln(x^2+9)=-\ln(x^2+9)$

Now, let differentiate

Recalled that the derivative of natural logs is exemplified as

$\ln(f(x))=\frac{f'(x)}{f(x)}$

$f(x)=x^2+9$, and the $f'(x)=2x$

$\frac{d}{dx}\left(-\ln(x^2+9)\right)=-(\frac{2x}{x^2+9})$

$\frac{d}{dx}\left(-\ln(x^2+9)\right)=-\frac{-2x}{x^2+9}$

That's all I've got for now. I recommend reading these posts if you want to learn more about differentiation:
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