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provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 76

Answers (1)

Answer: \frac{2}{\pi }

Hint: you must know the rules of solving derivative of trigonometric functions

Given: f(x)=\sqrt{\tan \sqrt{x}}


Find: f^{'}\left(\frac{\pi^{2}}{16}\right)
 

Solution:

f(x)=\sqrt{\tan (\sqrt{x})}

Differentiate with respect to x,

f^{\prime}(x)=\frac{1}{2 \sqrt{\tan (\sqrt{x})}} \frac{d}{d x} \tan \sqrt{x}

          =\frac{1}{2 \sqrt{\tan (\sqrt{x})}} \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}

f^{\prime}(x) \Rightarrow \frac{\sec ^{2} \sqrt{x}}{4 \sqrt{x \tan (\sqrt{x})}}

Now, f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{\sec ^{2} \sqrt{\frac{\pi^{2}}{16}}}{4 \sqrt{\frac{\pi^{2}}{16} \tan \left(\sqrt{\frac{\pi^{2}}{16}}\right)}}

                          =\frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{4\left(\frac{\pi}{4}\right) \sqrt{\tan \left(\frac{\pi}{4}\right)}}

                           \Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi \times(1)} \quad\left[B u t \tan \frac{\pi}{4} \quad \Rightarrow 1\right]

                           \begin{aligned} &\Rightarrow \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\pi} \\\\ &\Rightarrow \frac{\left(\sec \left(\frac{\pi}{4}\right)\right)^{2}}{\pi} \end{aligned}

                          \begin{aligned} &=\frac{(\sqrt{2})^{2}}{\pi} \\\\ &f^{\prime}\left(\frac{\pi^{2}}{16}\right)=\frac{2}{\pi} \end{aligned}
 

Posted by

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