Chapter 3, Problem 1RE

Find derivatives for the functions in Exercises 1–73. Assume a, b, c, and k are constants.

w = (t² + 1)¹⁰⁰

To determine

The derivative of the function $w={\left(t^2+1\right)}^{100}$.

Answer to Problem 1RE

The derivative of the function $w={\left(t^2+1\right)}^{100}$ is $\underset{\_}{200t{\left(t^2+1\right)}^{99}}$.

Explanation of Solution

Formula used:

The Chain Rule:

If f and g are differentiable functions, then $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$.

Derivative of Sum:

If f and g are differentiable functions, then $\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=f^{\prime }\left(x\right)+g^{\prime }\left(x\right)$.

The Power Rule:

For any constant real number n, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$.

Calculation:

Obtain the derivative of the function $w={\left(t^2+1\right)}^{100}$ as follows.

$\begin{array}{c}\frac{dw}{dt}=\frac{d}{dt}\left[{\left(t^2+1\right)}^{100}\right]\\ =100{\left(t^2+1\right)}^{100-1}\frac{d}{dt}\left(t^2+1\right)\left(\text{By Chain Rule}\right)\\ =100{\left(t^2+1\right)}^{99}\left(2t^{2-1}+0\right)\left(\text{By Derivative of Sum}\right)\\ =200t{\left(t^2+1\right)}^{99}\end{array}$

Therefore, the derivative of the function $w={\left(t^2+1\right)}^{100}$ is $\underset{\_}{200t{\left(t^2+1\right)}^{99}}$.