Q. 58 - Calculus

Question

Use the definition of the derivative to find the derivatives described in Exercises 55-58.

Given $\frac{d^{2}f}{d{x}^2}=2-x^2$, find $f^{\left(3\right)}\left(4\right)$ and $\frac{d^4}{d{x}^4}\left(f\right(x\left)\right)$

Step-by-Step Solution

Answer

The value is $-2$

Step 1. Given information

Given function $\frac{d^{2}f}{d{x}^2}=2-x^2$

Step 2: Use the definition of derivative

Calculating, we get

$$\begin{gathered} \underset{z\to x}{\lim }\frac{f^{\left(2\right)}\left(z\right)-f^{\left(2\right)}\left(x\right)}{z-x}=\underset{z\to x}{\lim }\frac{\left(2-z^2\right)-\left(2-x^2\right)}{z-x} \\ \underset{z\to x}{\lim }\frac{f^{\left(2\right)}\left(z\right)-f^{\left(2\right)}\left(x\right)}{z-x}=\underset{z\to x}{\lim }\frac{-\left(z^2-x^2\right)}{z-x} \\ =\underset{z\to x}{\lim }\frac{-(z-x)(z+x)}{z-x} \\ =\underset{z\to x}{\lim }-(z+x) \\ =-(x+x) \\ =-2x \\ {f}^{\left(3\right)}\left(4\right)=-2\left(4\right) \\ =-8 \\ \frac{d^4}{d{x}^4}\left(f\right(x\left)\right)=\underset{z\to x}{\lim }\frac{f^{\left(3\right)}\left(z\right)-f^{\left(3\right)}\left(x\right)}{z-x} \\ \frac{d^4}{d{x}^4}\left(f\right(x\left)\right)=\underset{z\to x}{\lim }\frac{(-2z)-(-2x)}{z-x} \\ =\underset{z\to x}{\lim }\frac{-2(z-x)}{z-x} \\ =\underset{z\to x}{\lim }-2 \\ =-2 \end{gathered}$$