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

    \[\begin{aligned} (f \circ g)^{\prime}(x) & =\frac{5}{2 \sqrt{4+\cos x}} \cdot(-\sin x) \\ = & \frac{-5 \sin x}{2 \sqrt{4+\cos x}} \end{aligned}\]