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Miscellaneous Examples · Example 41

Q.Find f′(x)f'(x) if f(x)=(sin⁡x)sin⁡xf(x) = (\sin x)^{\sin x} for all 0<x<π0 < x < \pi.

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Use logarithmic differentiation to handle a variable in both the base and exponent. Taking log⁡\log converts the product into a manageable form, then differentiate implicitly. The result is f′(x)=(sin⁡x)sin⁡x⋅cos⁡x⋅(1+log⁡(sin⁡x))f'(x) = (\sin x)^{\sin x} \cdot \cos x \cdot (1 + \log(\sin x)).

When you see a function where the variable appears in both the base and the exponent — like (sin⁡x)sin⁡x(\sin x)^{\sin x} — the standard power rule or exponential rule alone won't work. The power rule assumes a constant exponent; the exponential rule assumes a constant base. Here, both are moving.

The trick is to use logarithmic differentiation. Taking the natural logarithm transforms the exponent into a coefficient, letting you differentiate using the product rule. Then you solve for f′(x)f'(x) by multiplying back the original function.

Let’s walk through it.


  1. Set up the equation. Write y=f(x)=(sin⁡x)sin⁡xy = f(x) = (\sin x)^{\sin x}. Take the natural logarithm of both sides:

log⁡y=log⁡((sin⁡x)sin⁡x)\log y = \log\left( (\sin x)^{\sin x} \right)

Using the logarithm power rule: log⁡(ab)=blog⁡a\log(a^b) = b \log a, we get:

log⁡y=sin⁡x⋅log⁡(sin⁡x)\log y = \sin x \cdot \log(\sin x)

  1. Differentiate implicitly with respect to xx. On the left, the derivative of log⁡y\log y is 1y⋅y′\frac{1}{y} \cdot y' (by the chain rule). On the right, we have a product: sin⁡x\sin x times log⁡(sin⁡x)\log(\sin x). Use the product rule:

ddx[sin⁡x⋅log⁡(sin⁡x)]=(cos⁡x)⋅log⁡(sin⁡x)+sin⁡x⋅1sin⁡x⋅cos⁡x\frac{d}{dx} \big[ \sin x \cdot \log(\sin x) \big] = (\cos x) \cdot \log(\sin x) + \sin x \cdot \frac{1}{\sin x} \cdot \cos x

The second term simplifies: sin⁡x⋅cos⁡xsin⁡x=cos⁡x\sin x \cdot \frac{\cos x}{\sin x} = \cos x.

So the derivative of the right side is:

cos⁡x⋅log⁡(sin⁡x)+cos⁡x\cos x \cdot \log(\sin x) + \cos x

Factor out cos⁡x\cos x:

cos⁡x(log⁡(sin⁡x)+1)\cos x \big( \log(\sin x) + 1 \big)

Putting it together:

1y⋅y′=cos⁡x(1+log⁡(sin⁡x))\frac{1}{y} \cdot y' = \cos x \big( 1 + \log(\sin x) \big)

  1. Solve for y′y'. Multiply both sides by yy: …

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