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Find dy/DX when y=x^sinx+sin-1√x.
Please give me the solution of this problem.

bunti raj , 3 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

Last Activity: 5 Months ago

To find the derivative \( \frac{dy}{dx} \) for the function \( y = x^{\sin x} + \sin^{-1}(\sqrt{x}) \), we will differentiate each term separately using the rules of differentiation. Let's break it down step by step.

Step 1: Differentiate the first term \( x^{\sin x} \)

For the term \( x^{\sin x} \), we can use logarithmic differentiation. First, take the natural logarithm of both sides:

Let \( z = x^{\sin x} \). Then, taking the natural logarithm gives:

\( \ln z = \sin x \cdot \ln x \)

Now, differentiate both sides with respect to \( x \):

  • Using the chain rule on the left side: \( \frac{1}{z} \frac{dz}{dx} \)
  • For the right side, apply the product rule: \( \frac{d}{dx}(\sin x \cdot \ln x) = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \)

Putting it all together, we have:

\( \frac{1}{z} \frac{dz}{dx} = \cos x \cdot \ln x + \frac{\sin x}{x} \)

Now, multiply both sides by \( z \) (which is \( x^{\sin x} \)) to isolate \( \frac{dz}{dx} \):

\( \frac{dz}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \)

Step 2: Differentiate the second term \( \sin^{-1}(\sqrt{x}) \)

Next, we differentiate \( \sin^{-1}(\sqrt{x}) \). We will use the chain rule here:

  • The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \) where \( u = \sqrt{x} \).
  • Now, differentiate \( u = \sqrt{x} \) to get \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \).

Applying the chain rule gives us:

\( \frac{d}{dx}(\sin^{-1}(\sqrt{x})) = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} \)

Step 3: Combine the results

Now that we have both derivatives, we can combine them to find \( \frac{dy}{dx} \):

So, the derivative \( \frac{dy}{dx} \) is:

\( \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + \frac{1}{2\sqrt{x(1-x)}} \)

Final Expression

Thus, the final expression for the derivative is:

\( \frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + \frac{1}{2\sqrt{x(1-x)}} \)

This result gives you the rate of change of \( y \) with respect to \( x \) for the given function. If you have any further questions or need clarification on any steps, feel free to ask!

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