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

Answer: (B)

1

Answer 2

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.