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 the expression:

\[ y = x^{\sin(x)} + (\sin(x))^x \] 

We differentiate the expression with respect to \(x\) using the sum rule, which allows us to differentiate each term separately.

For the first term, \(x^{\sin(x)}\), the derivative is:

\[ \frac{d}{dx}(x^{\sin(x)}) = (\sin(x)) \cdot (x^{\sin(x)-1}) + (x^{\sin(x)}) \cdot (\ln(x)) \cdot \cos(x) \]

For the second term, \((\sin(x))^x\), the derivative is:

\[ \frac{d}{dx}((\sin(x))^x) = (\sin(x))^x \cdot \ln(\sin(x)) \cdot \cos(x) + (\sin(x))^{x-1} \cdot x \cdot \cos(x) \]

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 \(\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}}\), we substitute \(x = \frac{\pi}{2}\) into the equation:

\[ \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}} = (\sin(\frac{\pi}{2})) \cdot \left( \left(\frac{\pi}{2}\right)^{\sin(\frac{\pi}{2})-1} \right) + \left( \left(\frac{\pi}{2}\right)^1 \right) \cdot \left( \ln\left(\frac{\pi}{2}\right) \right) \cdot \cos(\frac{\pi}{2}) + (\sin(\frac{\pi}{2}))^{\frac{\pi}{2}} \cdot \ln(\sin(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) \]

Simplifying the trigonometric functions and evaluating the values:

\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = (1) \cdot \left( \left(\frac{\pi}{2}\right)^{1-1} \right) + \left( \left(\frac{\pi}{2}\right)^1 \right) \cdot \left( \ln\left(\frac{\pi}{2}\right) \right) \cdot (0) + (1)^{\frac{\pi}{2}} \cdot \ln(1) \cdot (0) + (1)^{\frac{\pi}{2}-1} \cdot \left( \frac{\pi}{2} \right) \cdot (0) \]

Substituting the values gives:

\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = (1) \cdot (1) + (1) \cdot \left( \ln\left(\frac{\pi}{2}\right) \right) \cdot (0) + (1) \cdot (0) + (1) \cdot \left( \frac{\pi}{2} \right) \cdot (0) \]

After simplifying:

\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = 1 + 0 + 0 + 0 \]

Thus, the value of \(\frac{dy}{dx}\) at \(x = \frac{\pi}{2}\) is 1, which corresponds to option (B) 1.

Answer 2

We are given: \[ y = x^{\sin x} + (\sin x)^x \] Let us differentiate both terms: 

1. Derivative of \(x^{\sin x}\): Use logarithmic differentiation: \[ \frac{d}{dx} (x^{\sin x}) = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right) \] 

2. Derivative of \((\sin x)^x\): Again, use logarithmic differentiation: \[ \frac{d}{dx} \left( (\sin x)^x \right) = (\sin x)^x \left( \log (\sin x) + x \cdot \cot x \right) \] Now evaluate at \(x = \frac{\pi}{2}\): - \(\sin\left(\frac{\pi}{2}\right) = 1\) - \(\cos\left(\frac{\pi}{2}\right) = 0\) - \(\log(1) = 0\) - \(\cot\left(\frac{\pi}{2}\right) = 0\) So, \[ x^{\sin x} = \left(\frac{\pi}{2}\right)^1 = \frac{\pi}{2} \] \[ \Rightarrow \frac{d}{dx}(x^{\sin x}) = \frac{\pi}{2} \cdot \left( \frac{1}{\frac{\pi}{2}} + 0 \right) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1 \] \[ (\sin x)^x = 1^x = 1 \] \[ \Rightarrow \frac{d}{dx}((\sin x)^x) = 1 \cdot (0 + x \cdot 0) = 0 \] 

Final answer: \[ \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}} = 1 + 0 = \mathbf{1} \]

Answer 3

Let \(y = x^{\sin(x)} + (\sin(x))^x\). We want to find \(\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}}\).

Let \(u = x^{\sin(x)}\) and \(v = (\sin(x))^x\). Then \(y = u + v\), so \(\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\).

First, let's find \(\frac{du}{dx}\):

Since \(u = x^{\sin(x)}\), take the natural logarithm of both sides:

\(\ln(u) = \sin(x) \ln(x)\)

Differentiate both sides with respect to x:

\(\frac{1}{u} \frac{du}{dx} = \cos(x) \ln(x) + \sin(x) \cdot \frac{1}{x}\)

\(\frac{du}{dx} = u \left[ \cos(x) \ln(x) + \frac{\sin(x)}{x} \right]\)

\(\frac{du}{dx} = x^{\sin(x)} \left[ \cos(x) \ln(x) + \frac{\sin(x)}{x} \right]\)

Next, let's find \(\frac{dv}{dx}\):

Since \(v = (\sin(x))^x\), take the natural logarithm of both sides:

\(\ln(v) = x \ln(\sin(x))\)

Differentiate both sides with respect to x:

\(\frac{1}{v} \frac{dv}{dx} = \ln(\sin(x)) + x \cdot \frac{\cos(x)}{\sin(x)}\)

\(\frac{dv}{dx} = v \left[ \ln(\sin(x)) + x \cot(x) \right]\)

\(\frac{dv}{dx} = (\sin(x))^x \left[ \ln(\sin(x)) + x \cot(x) \right]\)

Now, evaluate \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) at \(x = \frac{\pi}{2}\):

\(\frac{du}{dx} \Big|_{x = \frac{\pi}{2}} = \left(\frac{\pi}{2}\right)^{\sin(\frac{\pi}{2})} \left[ \cos(\frac{\pi}{2}) \ln\left(\frac{\pi}{2}\right) + \frac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} \right] = \left(\frac{\pi}{2}\right)^1 \left[ 0 \cdot \ln\left(\frac{\pi}{2}\right) + \frac{1}{\frac{\pi}{2}} \right] = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1\)

\(\frac{dv}{dx} \Big|_{x = \frac{\pi}{2}} = \left(\sin\left(\frac{\pi}{2}\right)\right)^{\frac{\pi}{2}} \left[ \ln\left(\sin\left(\frac{\pi}{2}\right)\right) + \frac{\pi}{2} \cot\left(\frac{\pi}{2}\right) \right] = (1)^{\frac{\pi}{2}} \left[ \ln(1) + \frac{\pi}{2} \cdot 0 \right] = 1 \cdot [0 + 0] = 0\)

Finally, \(\frac{dy}{dx} \Big|_{x = \frac{\pi}{2}} = \frac{du}{dx} \Big|_{x = \frac{\pi}{2}} + \frac{dv}{dx} \Big|_{x = \frac{\pi}{2}} = 1 + 0 = 1\)

Therefore, \(\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}} = 1\)

Answer: 1