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Chapter 7: Problem 31

Find the value of the derivative of the function at the given point. $$ f(x)=-\frac{1}{2} x\left(1+x^{2}\right) \quad(1,-1) $$

Short Answer

Expert verified
The value of the derivative \(f'(x)\) at the point \((1,-1)\) is -2.

Step by step solution

01

Differentiate the Function

The derivative of the function \(f(x)= -\frac{1}{2} x (1+x^2)\) is found by using both product and chain rule. Before differentiation, the function is rewritten like this: \(f(x)= -\frac{1}{2}x - \frac{1}{2}x^3\). Now differentiate both terms separately. The derivative of the first term is straightforward while the derivative of the second term is acquired by using the chain rule.\(f'(x)= -\frac{1}{2} - \frac{3}{2} x^2\).
02

Evaluate the derivative at the given point

The derivative \(f'(x)= -\frac{1}{2} - \frac{3}{2} x^2\) is then evaluated at the given point \((1,-1)\). So, substitute \(x=1\) into the derivative function, we get \(f'(1)= -\frac{1}{2} - \frac{3}{2} = -2\).

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