Exercise 5.4 Practice
Differentiate the following functions with respect to $x$
Q1
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Differentiate $\dfrac{e^x}{\sin x}$ with respect to $x$.
Step 1 (Let):
Let $y=\dfrac{e^x}{\sin x}$
Step 2 (Identify):
$u=e^x,\; v=\sin x$
Step 3 (Differentiate):
$u' = e^x,\; v'=\cos x$
Step 4 (Quotient Rule):
$\dfrac{dy}{dx}=\dfrac{vu'-uv'}{v^2}$
$\dfrac{dy}{dx}=\dfrac{\sin x\cdot e^x - e^x\cos x}{\sin^2x}$
$\boxed{\dfrac{dy}{dx}=\dfrac{e^x(\sin x-\cos x)}{\sin^2x}}$
Q2
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Differentiate $e^{\sin x}$ with respect to $x$.
Step 1:
Let $y=e^{\sin x}$
Step 2 (Chain Rule):
If $y=e^u$, then $\dfrac{dy}{dx}=e^u\dfrac{du}{dx}$
Step 3:
Here $u=\sin x \Rightarrow \dfrac{du}{dx}=\cos x$
$\boxed{\dfrac{dy}{dx}=e^{\sin x}\cos x}$
Q3
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Differentiate $e^{x^3}$ with respect to $x$.
Let $y=e^{x^3}$
Using chain rule
$\dfrac{dy}{dx}=e^{x^3}\cdot\dfrac{d}{dx}(x^3)$
$\boxed{\dfrac{dy}{dx}=3x^2e^{x^3}}$
Q4
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Differentiate $\sin(\tan^{-1}x)$ with respect to $x$.
Let $y=\sin(\tan^{-1}x)$
Using chain rule
$\dfrac{dy}{dx}=\cos(\tan^{-1}x)\cdot\dfrac{1}{1+x^2}$
$\boxed{\dfrac{dy}{dx}=\dfrac{\cos(\tan^{-1}x)}{1+x^2}}$
Q5
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Differentiate $\log(\cos x)$ with respect to $x$.
Let $y=\log(\cos x)$
Using $\dfrac{d}{dx}[\log u]=\dfrac{1}{u}\dfrac{du}{dx}$
$\dfrac{dy}{dx}=\dfrac{1}{\cos x}(-\sin x)$
$\boxed{\dfrac{dy}{dx}=-\tan x}$
Q6
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Differentiate $e^x+e^{2x}+e^{3x}$ with respect to $x$.
Differentiate term by term
$\dfrac{d}{dx}(e^x)=e^x$
$\dfrac{d}{dx}(e^{2x})=2e^{2x},\;
\dfrac{d}{dx}(e^{3x})=3e^{3x}$
$\boxed{\dfrac{dy}{dx}=e^x+2e^{2x}+3e^{3x}}$
Q7
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Differentiate $\sqrt{e^x},\;x>0$.
$y=\sqrt{e^x}=(e^x)^{1/2}$
$\dfrac{dy}{dx}=\dfrac{1}{2}(e^x)^{-1/2}\cdot e^x$
$\boxed{\dfrac{dy}{dx}=\dfrac{1}{2}\sqrt{e^x}}$
Q8
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Differentiate $\log(\log x),\;x>1$.
$y=\log(\log x)$
$\dfrac{dy}{dx}=\dfrac{1}{\log x}\cdot\dfrac{1}{x}$
$\boxed{\dfrac{dy}{dx}=\dfrac{1}{x\log x}}$
Q9
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Differentiate $\dfrac{\cos x}{\log x},\;x>0$.
Using quotient rule
$\dfrac{dy}{dx}=\dfrac{\log x(-\sin x)-\cos x\left(\dfrac{1}{x}\right)}{(\log x)^2}$
$\boxed{\dfrac{dy}{dx}=\dfrac{-\sin x\log x-\frac{\cos x}{x}}{(\log x)^2}}$
Q10
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Differentiate $\cos(\log x+e^x),\;x>0$.
Using chain rule
$\dfrac{dy}{dx}=-\sin(\log x+e^x)\left(\dfrac{1}{x}+e^x\right)$
$\boxed{\dfrac{dy}{dx}=-\sin(\log x+e^x)\left(\dfrac{1}{x}+e^x\right)}$