The given function is \( \displaystyle y = \sin \sqrt{1 – x^{2}} + x^{2} \cos 4x \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Step 1: Differentiate \( \sin \sqrt{1 – x^{2}} \)
Let \( u = \sqrt{1\, – x^{2}} = (1\, – x^{2})^{1/2} \), then
\(\displaystyle \frac{d}{du} (\sin u) = \cos u = \cos \sqrt{1 – x^{2}} \dots (1)\)
Now,
\(\displaystyle \frac{du}{dx} = \frac{1}{2}(1 – x^{2})^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 – x^{2}}} \dots (2) \)
By the chain rule (using (1) and (2)):
\( \displaystyle \frac{d}{dx} \left( \sin \sqrt{1 – x^{2}} \right) \\ \displaystyle = \cos \sqrt{1 – x^{2}} \cdot \left( \frac{-x}{\sqrt{1 – x^{2}}} \right) \\ \displaystyle = \frac{-x \cos \sqrt{1 – x^{2}}}{\sqrt{1 – x^{2}}} \\ \)Step 2: Differentiate \( x^{2} \cos 4x \)
Apply the product rule:
\( \displaystyle \frac{d}{dx} \left( x^{2} \cos 4x \right) \\ \displaystyle = 2x \cos 4x + x^{2} (-4 \sin 4x) \\ \displaystyle = 2x \cos 4x\, – 4x^{2} \sin 4x \\ \)Combine the results from step 1 and step 2:
\( \displaystyle \frac{dy}{dx} = \frac{d}{dx} \left( \sin \sqrt{1 – x^{2}} \right) + \frac{d}{dx} \left( x^{2} \cos 4x \right) \\ \displaystyle = \frac{-x \cos \sqrt{1 – x^{2}}}{\sqrt{1 – x^{2}}} + 2x \cos 4x\, – 4x^{2} \sin 4x \\ \)Therefore, \(\boxed{ \frac{dy}{dx} = \frac{-x \cos \sqrt{1 – x^{2}}}{\sqrt{1 – x^{2}}} + 2x \cos 4x\, – 4x^{2} \sin 4x } \)
Please let me know in the comments if you find any error in this solution.