Q 90 - Calculus

Question

Use implicit differentiation and the fact that $\frac{d}{dx}\left(x^4\right)=4x^3$ to prove that $\frac{d}{dx}\left(x^{-4}\right)=-4x^{-5}$. 

Step-by-Step Solution

Answer

$$\begin{gathered} y=x^{-4} \\ \Rightarrow y{x}^4=1 \\ \Rightarrow y\left[\frac{d}{dx}\left(x^4\right)\right]+x^4\frac{dy}{dx}=0 \\ \Rightarrow y\left(4x^3\right)+x^4\frac{dy}{dx}=0 \\ \Rightarrow 4x^{3}y+x^4\frac{dy}{dx}=0 \\ \Rightarrow x^4\frac{dy}{dx}=-4x^{3}y \\ \Rightarrow \frac{dy}{dx}=-\frac{4x^{3}y}{x^4} \\ \Rightarrow \frac{dy}{dx}=-\frac{4y}{x} \\ \Rightarrow \frac{d}{dx}\left(x^{-4}\right)=-\frac{4x^{-4}}{x} \\ \Rightarrow \frac{d}{dx}\left(x^{-4}\right)=-4x^{-4}{x}^{-1} \\ \Rightarrow \frac{d}{dx}\left(x^{-4}\right)=-4x^{-5} \end{gathered}$$

Hence proved.

Step 1. Given Information

We have given that :-

$\frac{d}{dx}\left(x^{-4}\right)=-4x^{-5}$.

We have to prove this derivative by using implicit differentiation and the following derivative :-

$\frac{d}{dx}\left(x^4\right)=4x^3$.

Step 2. Prove that d d x ( x - 4 ) = - 4 x - 5

We have to find the derivative of $x^{-4}$.

Let :-

$y=x^{-4}$.

We can write it as :-

$y{x}^4=1$.

Then by using implicit differentiation and product rule, we have :-

$y\left[\frac{d}{dx}\left(x^4\right)\right]+x^4\frac{dy}{dx}=0$

We have given that :-

$\frac{d}{dx}\left(x^4\right)=4x^3$. By using this we have :-

$$\begin{gathered} y\left(4x^3\right)+x^4\frac{dy}{dx}=0 \\ \Rightarrow 4x^{3}y+x^4\frac{dy}{dx}=0 \\ \Rightarrow x^4\frac{dy}{dx}=-4x^{3}y \\ \Rightarrow \frac{dy}{dx}=-\frac{4x^{3}y}{x^4} \\ \Rightarrow \frac{dy}{dx}=-\frac{4y}{x} \end{gathered}$$

Put the value $y=x^{-4}$, then we have :-

$$\begin{gathered} \frac{d}{dx}\left(x^{-4}\right)=-\frac{4x^{-4}}{x} \\ \Rightarrow \frac{d}{dx}\left(x^{-4}\right)=-4x^{-4}{x}^{-1} \\ \Rightarrow \frac{d}{dx}\left(x^{-4}\right)=-4x^{-5} \end{gathered}$$

This is the required value.

Hence proved.