Question

Let

Hint:

For differentiation of an integral with variable upper limit, use the Fundamental Theorem of Calculus along with the chain rule.

Answer 1

Correct Answer: 0.4

Step 1: Write the outer integral form.
Let
\[ F(y)=\int_0^{\log_e y} e^{-\frac{1}{\sqrt{2}}}(1-t^2)\,dt \]
Then
\[ f(x)=\int_0^{e^{\sin x}}F(y)\,dy \]

Step 2: Differentiate using Fundamental Theorem of Calculus.
\[ f'(x)=F(e^{\sin x})\cdot \frac{d}{dx}(e^{\sin x}) \]

Step 3: Differentiate upper limit.
\[ \frac{d}{dx}(e^{\sin x})=e^{\sin x}\cos x \]

Step 4: Evaluate at \(x=\frac{\pi}{4}\).
\[ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}, \qquad \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} \]
So,
\[ \log_e(e^{\sin x})=\sin x=\frac{1}{\sqrt{2}} \]

Step 5: Evaluate \(F(e^{\sin x})\).
\[ F(e^{\sin x}) = \int_0^{\frac{1}{\sqrt{2}}}e^{-\frac{1}{\sqrt{2}}}(1-t^2)\,dt \]
\[ = e^{-\frac{1}{\sqrt{2}}}\left[t-\frac{t^3}{3}\right]_0^{\frac{1}{\sqrt{2}}} \]

Step 6: Substitute the limits.
\[ F(e^{\sin x}) = e^{-\frac{1}{\sqrt{2}}} \left(\frac{1}{\sqrt{2}}-\frac{1}{3(2\sqrt{2})}\right) \]
\[ = e^{-\frac{1}{\sqrt{2}}}\cdot \frac{5}{6\sqrt{2}} \]

Step 7: Final calculation.
\[ f'\left(\frac{\pi}{4}\right) = e^{-\frac{1}{\sqrt{2}}}\cdot \frac{5}{6\sqrt{2}} \cdot e^{\frac{1}{\sqrt{2}}}\cdot \frac{1}{\sqrt{2}} \]
\[ = \frac{5}{12} = 0.4166\ldots \]
Rounded off to one decimal place,
\[ \boxed{0.4} \]