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Q4  Differentiate w.r.t. x the function in Exercises 1 to 11.

       \sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1

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Given function is
f(x)=\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^{-1}x\sqrt x)}{dx}=\frac{1}{\sqrt{1-(x\sqrt x)^2}}.\frac{d(x\sqrt x)}{dx}
                                                                         =\frac{1}{\sqrt{1-x^3}}.\left ( 1.\sqrt x+x\frac{1}{2\sqrt x} \right )
                                                                         =\frac{1}{\sqrt{1-x^3}}.\left ( \frac{3\sqrt x}{2} \right )
                                                                         =\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}

Therefore, differentiation w.r.t. x is   \frac{3}{2}.\sqrt{\frac{x}{1-x^3}}

Posted by

Gautam harsolia

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