Question

What is the derivative of ${{\left( 2x-3 \right)}^{5}}$?

Answer 1

Assume the function $\left( 2x-3 \right)$ as $f\left( x \right)$ and write ${{\left( 2x-3 \right)}^{5}}$ as ${{\left[ f\left( x \right) \right]}^{5}}$. Now, use the chain rule of differentiation to differentiate the function. First differentiate the function ${{\left[ f\left( x \right) \right]}^{5}}$ with respect to the function $f\left( x \right)$ and then differentiate the function $f\left( x \right)$ with respect to x. Finally, take the product of these two derivatives to get the answer. Use the formulas $\dfrac{d\left[ {{\left( f\left( x \right) \right)}^{n}} \right]}{d\left[ f\left( x \right) \right]}=n{{\left( f\left( x \right) \right)}^{n-1}}$ and $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ to get the answer. Use the fact that the derivative of a constant term is 0 to differentiate the terms inside the bracket.

Complete step by step answer:
Here we have been provided with the function ${{\left( 2x-3 \right)}^{5}}$ and we are asked to find its derivative. Here we will use the chain rule of derivatives to get the answer. Assuming the function $\left( 2x-3 \right)$ as $f\left( x \right)$ we have the function ${{\left( 2x-3 \right)}^{5}}$ of the form ${{\left[ f\left( x \right) \right]}^{5}}$. So we have,
$\Rightarrow {{\left( 2x-3 \right)}^{5}}={{\left[ f\left( x \right) \right]}^{5}}$
On differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=\dfrac{d{{\left[ f\left( x \right) \right]}^{5}}}{dx}$
Now, according to the chain rule of derivative first we have to differentiate the function ${{\left[ f\left( x \right) \right]}^{5}}$ with respect to $f\left( x \right)$ and then we have to differentiate $f\left( x \right)$ with respect to x. Finally, we need to consider their product to get the relation. So we get,
$\Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=\dfrac{d{{\left[ f\left( x \right) \right]}^{5}}}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$
Using the formula $\dfrac{d\left[ {{\left( f\left( x \right) \right)}^{n}} \right]}{d\left[ f\left( x \right) \right]}=n{{\left( f\left( x \right) \right)}^{n-1}}$ we get,
$\begin{aligned} & \Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=5{{\left[ f\left( x \right) \right]}^{5-1}}\times \dfrac{d\left[ f\left( x \right) \right]}{dx} \\\ & \Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=5{{\left[ f\left( x \right) \right]}^{4}}\times \dfrac{d\left[ f\left( x \right) \right]}{dx} \\\ \end{aligned}$
Substituting the value of $f\left( x \right)$ we get,
$\begin{aligned} & \Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=5{{\left[ 2x-3 \right]}^{4}}\times \dfrac{d\left[ 2x-3 \right]}{dx} \\\ & \Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=5{{\left[ 2x-3 \right]}^{4}}\times \left[ \dfrac{d\left[ 2x \right]}{dx}-\dfrac{d\left[ 3 \right]}{dx} \right] \\\ \end{aligned}$
Now, since 2 is a constant multiplied to the variable x so it can be taken out of the derivative. However, 3 is a constant term so its derivative will be 0, therefore we get,
$\begin{aligned} & \Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=5{{\left[ 2x-3 \right]}^{4}}\times \left[ 2\times \dfrac{d\left[ x \right]}{dx}-0 \right] \\\ & \Rightarrow \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=5{{\left[ 2x-3 \right]}^{4}}\times \left[ 2 \right] \\\ & \therefore \dfrac{d\left[ {{\left( 2x-3 \right)}^{5}} \right]}{dx}=10{{\left( 2x-3 \right)}^{4}} \\\ \end{aligned}$
Hence, the above relation is our answer.

Note: You must remember all the basic rules and formulas of differentiation like: - product rule, chain rule, $\dfrac{u}{v}$ rule etc. You may note an important formula from the above solution that is we know that the derivative of ${{x}^{n}}$ is $n{{x}^{n-1}}$ so if we take a function of the form${{\left( ax+b \right)}^{n}}$, i.e. linear in x where a and b are constant, then we will get the derivative formula as $\dfrac{d{{\left( ax+b \right)}^{n}}}{dx}=a{{\left( ax+b \right)}^{n-1}}$. This is a result of the chain rule you may remember to solve the question in less time.