Calculus Solutions Ex#2.6

 

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Q.01
Solution:
Given that $f^{\prime}(0)=2, g(0)=0$ and $g^{\prime}(0)=3$  find  $(f \circ g)^{\prime}(2).$  $$ \begin{aligned}  \text {Formula} (f \circ g)^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x) \quad \text { Put } x=0, \end{aligned}$$  $$ \begin{aligned}(f \circ g)^{\prime}(0) & =f^{\prime}(g(0)) g^{\prime}(0) & & \\ \\ & =f^{\prime}(0) \cdot(3) & \therefore g(0) =0 \\ \\ & =2(3)=6 & & g^{\prime}(0)=3 \\ \\ & & & f^{\prime}(0)=2 \end{aligned} $$
Q.02
 
 
Solution:
Given that $f^{\prime}(9)=5, g(2)=9$ and $g^{\prime}(2)=-3$ find $(f \circ g)^{\prime}(2)$. Formula $(f \circ g)^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x)$ $$ \text { Put } x=2, \begin{aligned} (f \circ g)^{\prime}(2) & =f^{\prime}(g(2)) g^{\prime}(2) & & \\ \\ & =f^{\prime}(9) \cdot(-3) & \therefore g(2) & =9 \\ \\ & =5(-3)=-15 & & g^{\prime}(2)=-3 \\ \\ & & & f^{\prime}(9)=5 \end{aligned} $$
Q.03 Let $f(x)=x^5$ and $g(x)=2 x-3$ (a) find $(f \circ g)(x)$ and $(f \circ g)^{\prime}(x)$
Solution:
$$ \begin{array}{rlrl} (f \circ g)(x)=f(g(x)) & =[g(x)]^5 & & \\ \\ & =(2 x-3)^5 & \therefore g(x)=x^5 \\ \\ & & \end{array} $$ and $$(f \circ g)^{\prime}(x)=f^{\prime}(g(x)) g^{\prime}(x)-(1)$$ 

Since $$g(x)=2 x-3$$ 
Differentiating w.r.t. $x$ $$ g^{\prime}(x)=2-(2) $$ and $$f(x)=x^5$$
Differentiating w.r.t $x$ $$ \begin{aligned} & f^{\prime}(x)=5 x^4 \\ \\ & f^{\prime}(g(x))=5(g(x))^4=5(2 x-3)^4-(3) \end{aligned} $$ Substituting (2) and (3) into (1)

$$ (f \circ g)^{\prime}(x)=5(2 x-3)^4(2)=10(2 x-3)^4 $$ (b) Find $(g \circ f)(x)$ and $(g \circ f)^{\prime}(x)$.

$$ \begin{array}{rlrl} (g \circ f)(x)=g(f(x)) & =2 f(x)-3 & \therefore g(x) & =2 x-3 \\ \\ & =2 x^5-3 & \therefore f(x)=x^5 \end{array} $$ $$ (g \circ f)^{\prime}(x)=g^{\prime}(f(x)) f^{\prime}(x)-(1) $$  Since $$g(x)=2 x-3$$ Differentiating w.r.t $x$ $$ g^{\prime}(x)=2-(2) $$ $$ \Rightarrow \quad g^{\prime}(f(x))=2 $$ and $$f(x)=x^5$$ Differentiating w.r.t $x$ $$ f^{\prime}(x)=5 x^{5-1}=5 x^4-(3) $$ Substituting (2) and (3) into (1)

$$ (g \circ f)^{\prime}(x)=2\left(5 x^4\right)=10 x^4 $$
Q.04 Let $f(x)=5 \sqrt{x}$ and $g(x)=4+\cos x$ (a) Find $(f \circ g)(x)$ and $(f \circ g)^{\prime}(x)$
Solution:
Solution $$ \begin{array}{rlrl} (f \circ g)(x) & =f(g(x))=5 \sqrt{g(x)} & & \therefore f(x)=5 \sqrt{x} \\ \\ & =5 \sqrt{4+ \cos x} & & \therefore g(x)=4+\cos x \\ \\ (f \circ g)^{\prime}(x) & =f^{\prime}(g(x)) g^{\prime}(x)-(1) \end{array} $$ Since $$f(x)=5 \sqrt{x}=5 x^{1 / 2}$$

Differentiating w.r.t $x$ $$ \begin{aligned} f^{\prime}(x) & =5\left(\frac{1}{2} x^{\frac{1}{2}-1}\right)=\frac{5}{2} x^{-1 / 2} \\ \\ f^{\prime}(x) & =\frac{5}{2 \sqrt{x}} \\ \\ \Rightarrow \quad f^{\prime}(g(x)) & =\frac{5}{2 \sqrt{g(x)}} \\ \\ f^{\prime}(g(x)) & =\frac{5}{2 \sqrt{4+\cos x}}-(2) \quad \therefore g(x)=4+\cos x \\ \\ \text { and since} \quad g(x) & =4+\cos x \end{aligned} $$ differentiating w.t.t $x$ $$ \begin{aligned} & g^{\prime}(x)=\frac{d}{d x}[4+\cos x] \\ \\ & g^{\prime}(x)=\frac{d}{d x}[4]+\frac{d}{d x}[\cos x] \\ \\ & g^{\prime}(x)=0-\sin x =-\sin x -(3) \end{aligned} $$ Substituting (2) and (3) into (1) we get $$ \begin{aligned} (f \circ g)^{\prime}(x) & =\frac{5}{2 \sqrt{4+\cos x}} \cdot(-\sin x) \\ = & \frac{-5 \sin x}{2 \sqrt{4+\cos x}} \end{aligned} $$