Exercise 5.2 Practice
Differentiate the following functions with respect to $x$
Q1
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Differentiate $\sin(x^3 + 4)$ with respect to $x$.
Given: Let $y = \sin(x^3 + 4)$
Concept: By Chain Rule, if $y = f(g(x))$, then
$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\sin(x^3 + 4)]$
$\frac{dy}{dx} = \frac{d}{dx} [\sin(x^3 + 4)]$
$= \cos(x^3 + 4) \cdot \frac{d}{dx}(x^3 + 4)$
$= \cos(x^3 + 4) \cdot (3x^2 + 0)$
$= \cos(x^3 + 4) \cdot (3x^2 + 0)$
$\frac{dy}{dx} = 3x^2 \cos(x^3 + 4)$
Q2
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Differentiate $\cos(\sin(x^2))$ with respect to $x$.
Step 1: Apply outer derivative of cosine.
$\frac{dy}{dx} = -\sin(\sin(x^2)) \cdot \frac{d}{dx}[\sin(x^2)]$
$\frac{dy}{dx} = -\sin(\sin(x^2)) \cdot \frac{d}{dx}[\sin(x^2)]$
Step 2: Apply derivative of inner sine
function.
$= -\sin(\sin(x^2)) \cdot \cos(x^2) \cdot \frac{d}{dx}(x^2)$
$= -\sin(\sin(x^2)) \cdot \cos(x^2) \cdot \frac{d}{dx}(x^2)$
$\frac{dy}{dx} = -2x \cdot \cos(x^2) \cdot \sin(\sin(x^2))$
Q3
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Differentiate $\sin(ax^2 + b)$ with respect to $x$.
$\frac{dy}{dx} = \cos(ax^2 + b) \cdot \frac{d}{dx}(ax^2 + b)$
$\frac{dy}{dx} = 2ax \cos(ax^2 + b)$
Q4
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Differentiate $\sec(\sin(\sqrt{x}))$ with respect to $x$.
$\frac{dy}{dx} = \sec(\sin\sqrt{x}) \tan(\sin\sqrt{x}) \cdot \frac{d}{dx}(\sin\sqrt{x})$
$= \sec(\sin\sqrt{x}) \tan(\sin\sqrt{x}) \cdot \cos\sqrt{x} \cdot \frac{d}{dx}(\sqrt{x})$
$= \sec(\sin\sqrt{x}) \tan(\sin\sqrt{x}) \cdot \cos\sqrt{x} \cdot \frac{1}{2\sqrt{x}}$
$= \sec(\sin\sqrt{x}) \tan(\sin\sqrt{x}) \cdot \cos\sqrt{x} \cdot \frac{d}{dx}(\sqrt{x})$
$= \sec(\sin\sqrt{x}) \tan(\sin\sqrt{x}) \cdot \cos\sqrt{x} \cdot \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx} = \frac{\cos\sqrt{x} \sec(\sin\sqrt{x})
\tan(\sin\sqrt{x})}{2\sqrt{x}}$
Q5
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Differentiate $\frac{\sin(ax+b)}{\cos(px+q)}$ with respect to $x$.
Concept: Quotient Rule $(\frac{u}{v})' = \frac{vu'
- uv'}{v^2}$
$u = \sin(ax+b) \Rightarrow u' = a\cos(ax+b)$
$v = \cos(px+q) \Rightarrow v' = -p\sin(px+q)$
$v = \cos(px+q) \Rightarrow v' = -p\sin(px+q)$
$\frac{dy}{dx} = \frac{\cos(px+q)[a\cos(ax+b)] - \sin(ax+b)[-p\sin(px+q)]}{\cos^2(px+q)}$
$\frac{dy}{dx} = \frac{a\cos(px+q)\cos(ax+b) +
p\sin(ax+b)\sin(px+q)}{\cos^2(px+q)}$
Q6
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Differentiate $\cos(x^4) \cdot \sin^2(x^5)$ with respect to $x$.
Concept: Product Rule $(uv)' = u'v + uv'$
$\frac{dy}{dx} = \cos(x^4) \cdot [2\sin(x^5)\cos(x^5) \cdot 5x^4] + \sin^2(x^5) \cdot [-\sin(x^4)
\cdot 4x^3]$
$\frac{dy}{dx} = 10x^4 \sin(x^5)\cos(x^5)\cos(x^4) - 4x^3 \sin^2(x^5)\sin(x^4)$
Q7
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Differentiate $3\sqrt{\cot(x^4)}$ with respect to $x$.
$\frac{dy}{dx} = 3 \cdot \frac{1}{2\sqrt{\cot(x^4)}} \cdot [-\csc^2(x^4)] \cdot 4x^3$
$\frac{dy}{dx} = \frac{-6x^3 \csc^2(x^4)}{\sqrt{\cot(x^4)}}$
Q8
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Differentiate $\cos(\sqrt{x^2+1})$ with respect to $x$.
$\frac{dy}{dx} = -\sin(\sqrt{x^2+1}) \cdot \frac{d}{dx}(\sqrt{x^2+1})$
$= -\sin(\sqrt{x^2+1}) \cdot \frac{1}{2\sqrt{x^2+1}} \cdot 2x$
$= -\sin(\sqrt{x^2+1}) \cdot \frac{1}{2\sqrt{x^2+1}} \cdot 2x$
$\frac{dy}{dx} = \frac{-x \sin(\sqrt{x^2+1})}{\sqrt{x^2+1}}$
Q9: Proof
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Prove that the function $f$ given by $f(x) = |x - 2|, x \in \mathbb{R}$ is not
differentiable at $x = 2$.
Concept: Differentiability requires $LHD = RHD$
Left Hand Derivative (LHD) at 2:
$\lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0} \frac{|(2+h)-2| - |2-2|}{h}$
$= \lim_{h \to 0} \frac{|h|}{h}$. As $h < 0, |h|=-h$.
$= \lim_{h \to 0} \frac{-h}{h} = -1$.
$\lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0} \frac{|(2+h)-2| - |2-2|}{h}$
$= \lim_{h \to 0} \frac{|h|}{h}$. As $h < 0, |h|=-h$.
$= \lim_{h \to 0} \frac{-h}{h} = -1$.
Right Hand Derivative (RHD) at 2:
$\lim_{h \to 0^+} \frac{|h|}{h}$. As $h > 0, |h| = h$.
$= \lim_{h \to 0} \frac{h}{h} = 1$.
$\lim_{h \to 0^+} \frac{|h|}{h}$. As $h > 0, |h| = h$.
$= \lim_{h \to 0} \frac{h}{h} = 1$.
Since $LHD (-1) \neq RHD (1)$, the function is not differentiable at $x = 2$.
Q10: Proof
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Prove that the greatest integer function defined by $f(x) = [x], 0 < x < 4$ is
not differentiable at $x=1, 2,$ and $3$.
General Check (at x=c where c is integer):
LHD: $\lim_{h \to 0^-} \frac{[c+h]-[c]}{h} = \frac{(c-1)-c}{h} =
\frac{-1}{h} \rightarrow \infty$
RHD: $\lim_{h \to 0^+} \frac{[c+h]-[c]}{h} = \frac{c-c}{h} = 0$.
RHD: $\lim_{h \to 0^+} \frac{[c+h]-[c]}{h} = \frac{c-c}{h} = 0$.
The derivative does not exist (infinite/mismatch) at integral points.
Hence, not differentiable at $x=1, 2, 3$.