Question

If \(y = x^{\sin(x)} + (\sin(x))^x\), then \(\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}}\) is

• A. \(\frac{4}{\pi}\)
• B. 1
• C. \(\pi \log \left(\frac{\pi}{2}\right)\)
• D. \(\frac{\pi^2}{2}\)

Answer 1

The Correct Option is B

Given expression \(y = x^{\sin(x)} + (\sin(x))^x\) with respect to x.
Using the sum rule, we differentiate each term separately.
For the first term, \(x^{\sin(x)}\)
\(\frac{d}{dx}(x^{\sin(x)}) = (\sin(x)) \cdot (x^{\sin(x)-1}) + (x^{\sin(x)}) \cdot (\ln(x)) \cdot \cos(x)\)

For the second term, \((\sin(x))^x\)
\(\frac{d}{dx}((\sin(x))^x) = (\sin(x))^x \cdot \ln(\sin(x)) \cdot \cos(x) + (\sin(x))^{x-1} \cdot x \cdot \cos(x)\)

Now, we can find \(\frac{dy}{dx}\) by adding the derivatives of both terms:
\(\frac{dy}{dx} = (\sin(x)) \cdot (x^{\sin(x)-1}) + (x^{\sin(x)}) \cdot (\ln(x)) \cdot \cos(x) + (\sin(x))^x \cdot \ln(\sin(x)) \cdot \cos(x) + (\sin(x))^{x-1} \cdot x \cdot \cos(x)\)

To evaluate \(\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}} = (\sin(\frac{\pi}{2})) \cdot (\frac{\pi}{2}^{\sin(\frac{\pi}{2})-1}) + (\frac{\pi}{2}^{\sin(\frac{\pi}{2})}) \cdot (\ln(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) + (\sin(\frac{\pi}{2}))^{\frac{\pi}{2}} \cdot \ln(\sin(\frac{\pi}{2})\)
Simplifying the trigonometric functions and evaluating the values:
\(\frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = (1) \cdot \left( \left(\frac{\pi}{2}\right)^{1-1} \right) + \left( \left(\frac{\pi}{2}\right)^1 \right) \cdot \left( \ln\left(\frac{\pi}{2}\right) \right) \cdot (0) + (1)^{\frac{\pi}{2}} \cdot \ln(1) \cdot (0) + (1)^{\frac{\pi}{2}-1} \cdot \left( \frac{\pi}{2} \right) \cdot (0)\)
\(\frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = (1) \cdot (1) + (1) \cdot \left( \ln\left(\frac{\pi}{2}\right) \right) \cdot (0) + (1) \cdot (0) + (1) \cdot \left( \frac{\pi}{2} \right) \cdot (0\)
\(\frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = 1 + 0 + 0 + 0\)
\(\frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = 1\)
Therefore, the value of \(\frac{dy}{dx} \text{ at } x = \frac{\pi}{2} \text{ is } 1\) corresponding to option (B) 1.