Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative of Function and absolute value.

## Derivative of Function and absolute value (B.Stat Objective Question )

Let \(f(x)=|sin^{3}x|\), \(g(x)=sin^{3}x\), both being defined for x in the interval \((\frac{-\pi}{2}, \frac{\pi}{2})\). Then

- \(f'(x)=g'(x) for all x\)
- \(g'(x)=|f'(x)| for all x\)
- \(f'(x)=|g'(x)| for all x\)
- \(f'(x)=-g'(x) for all x\)

**Key Concepts**

Equation

Derivative

Algebra

## Check the Answer

But try the problem first…

Answer:\(g'(x)=|f'(x)| for all x\)

B.Stat Objective Problem 767

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Let \(f(x) =|sin^{3}x|=sin^{2}x|sinx|\)

\(f'(x)=2cos(x)sin(x)|sin(x)|+\frac{cos(x)(sin^{3}x)}{|sin(x)|}\)

=\(3sin(x)cos(x)|sin(x)|\)

Second Hint

or, \(g(x)=sin^{3}x\)

or, \(g'(x)=3 cosx sin^{2}x\)

Final Step

or, |f'(x)| for all \(x \in (\frac{-\pi}{2},\frac{\pi}{2})\) = \(3cos x|sin x||sin x|=g'(x)\)

or, |f'(x)|=g'(x) for all \(x \in (\frac{-\pi}{2},\frac{\pi}{2})\)

Google