Chapter 4.1, Problem 35E

To determine

To calculate: $(f.g)\text{'}$ for the given function at given value of x.

Answer to Problem 35E

The answer is $\frac{d(f(g(1)))}{dx}=-\frac{\pi }{4}$ .

Explanation of Solution

Given information: Given functions are $f(u)=\cot \frac{\pi u}{10},u=g(x)=5\sqrt{x}\text{and}\text{value}\text{of}x=1$ .

Formula used:

  $\begin{array}{l}\frac{d(f(g(x)))}{dx}=f\text{'}(g(x))\times g\text{'}(x)\\ \frac{d(\sqrt{x})}{dx}=\frac{1}{2\sqrt{x}}\\ \frac{d(\cot x)}{dx}=-\cos e{c}^{2}x\end{array}$

Also, quotient, chain and addition rule.

Calculation:

If $f(u)=\cot \frac{\pi u}{10},u=g(x)=5\sqrt{x}$

Then,

  $f(g(x))=\cot \frac{\pi (5\sqrt{x})}{10}$

Differentiating the function w.r.t x we get,

  $\begin{array}{c}\frac{d(f(g(x)))}{dx}=\frac{d\left(\cot \frac{\pi (5\sqrt{x})}{10}\right)}{dx}\\ =\left(-\cos e{c}^2\left(\frac{\pi (5\sqrt{x})}{10}\right)\times \frac{\pi }{10}\right)\times \frac{5}{2\sqrt{x}}\\ =\frac{-5\pi }{20}\times \cos e{c}^2\left(\frac{\pi \left(5\sqrt{x}\right)}{10}\right)\times \frac{1}{\sqrt{x}}\\ =\frac{-\pi }{4}\times \cos e{c}^2\left(\frac{\pi \left(\sqrt{x}\right)}{2}\right)\times \frac{1}{\sqrt{x}}\end{array}$

Now $x=1$ in the above equation we get,

  $\begin{array}{c}\frac{d(f(g(1)))}{dx}=\frac{-\pi }{4}\times \cos e{c}^2\left(\frac{\pi \left(5\sqrt{(1)}\right)}{10}\right)\times \frac{1}{\sqrt{(1)}}\\ =-\frac{\pi }{4}\times \cos e{c}^2\left(\frac{\pi \times (1)}{2}\right)\times \frac{1}{(1)}\\ =-\frac{\pi }{4}\times \cos e{c}^2\left(\frac{\pi }{2}\right)\\ =-\frac{\pi }{4}\times {(}^1=-\frac{\pi }{4}\end{array}$

The answer is $\frac{d(f(g(1)))}{dx}=-\frac{\pi }{4}$ .