Question

Let $f(x)={{x}^{5}}$ and $g(x)=2x-3$. Find $\left( fog \right)(x)$ and $\left( fog \right)'(x)$.

Answer 1

Hint:To find $\left( fog \right)(x)$, substitute (2x – 3) in place of ‘x’ in the function $f(x)={{x}^{5}}$. Now, to find $\left( fog \right)'(x)$, differentiate the function obtained by substituting (2x – 3) in place of ‘x’ in the function $f(x)={{x}^{5}}$. Use the formula: $\dfrac{d}{dx}{{\left[ F\left( x \right) \right]}^{n}}=n{{\left[ F\left( x \right) \right]}^{n-1}}\dfrac{d}{dx}\left[ F\left( x \right) \right]$, where $F\left( x \right)$ is the function obtained after the first step, to get the required derivative.

Complete step-by-step answer:
We have been provided with two functions: $f(x)={{x}^{5}}$ and $g(x)=2x-3$, and we have to find $\left( fog \right)(x)$ and $\left( fog \right)'(x)$.
Here, $\left( fog \right)(x)$ is called a composite function. In simplified form it can be written as:
$\left( fog \right)(x)=f\left( g(x) \right)$
To find the value of $\left( fog \right)(x)$, we have to substitute the value of $g(x)$ in place of ‘x’, in the function $f(x)$.
Therefore, in the above question, we have to substitute (2x – 3) in place of ‘x’ in the function $f(x)={{x}^{5}}$.
Substituting (2x – 3) in place of ‘x’ in the function $f(x)={{x}^{5}}$, we get,
$\begin{aligned} & \left( fog \right)(x)=f\left( g(x) \right) \\\ & ={{\left( 2x-3 \right)}^{5}} \\\ \end{aligned}$
Now, to find $\left( fog \right)'(x)$, we must differentiate $\left( fog \right)(x)$.
Since, $\left( fog \right)(x)={{\left( 2x-3 \right)}^{5}}$ is of the form ${{\left[ F\left( x \right) \right]}^{n}}$, where $F\left( x \right)=\left( 2x-3 \right)$ and n = 5. Therefore, using the formula: $\dfrac{d}{dx}{{\left[ F\left( x \right) \right]}^{n}}=n{{\left[ F\left( x \right) \right]}^{n-1}}\dfrac{d}{dx}\left[ F\left( x \right) \right]$, we get,

& {{\left( fog \right)}^{'}}\left( x \right)=\dfrac{d}{dx}{{\left[ 2x-3 \right]}^{5}} \\\ & =5\times {{\left[ 2x-3 \right]}^{4}}\times \dfrac{d}{dx}\left[ 2x-3 \right] \\\ & =5\times {{\left[ 2x-3 \right]}^{4}}\times 2 \\\ & =10\times {{\left[ 2x-3 \right]}^{4}} \\\ & =10{{\left[ 2x-3 \right]}^{4}} \\\ \end{aligned}$$ Note: One may note that we can also break the terms of ${{\left( 2x-3 \right)}^{5}}$ by using the binomial formula: ${{\left( 2x-3 \right)}^{5}}=\sum\limits_{r=1}^{5}{{}^{5}{{C}_{r}}}{{\left( 2x \right)}^{5-r}}{{\left( -3 \right)}^{r}}$. We will get a total of six terms. Then we can differentiate these terms one by one. But this will be a lengthy process and the chances of making calculation mistakes are more. You may note that the answer will not change.