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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

But try the problem first…

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

Source

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 $x \in (\frac{-\pi}{2},\frac{\pi}{2})$