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Exercise 5.7 · Q10

Q.Find dydx\frac{dy}{dx} in the following: sin⁡(log⁡x)\sin (\log x)

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The derivative of sin⁡(log⁡x)\sin(\log x) is found by applying the Chain Rule: differentiate the outer sine function, then multiply by the derivative of the inner log⁡x\log x. The result is dydx=cos⁡(log⁡x)x\frac{dy}{dx} = \frac{\cos(\log x)}{x}.

The key idea here is the Chain Rule. Whenever you have a function of a function — like sin⁡\sin of log⁡x\log x — you cannot differentiate it in one step. The sine function “sees” log⁡x\log x as its input, not xx directly. So you first differentiate the outer function (sine) with respect to its own input, and then multiply by the derivative of that inner input (log⁡x\log x) with respect to xx.

Let’s walk through it carefully.

  1. Identify the outer and inner functions.

    We have y=sin⁡(log⁡x)y = \sin(\log x).

    The outer function is sin⁡(u)\sin(u), where u=log⁡xu = \log x is the inner function.

  2. Differentiate the outer function with respect to its inner input.

    The derivative of sin⁡(u)\sin(u) with respect to uu is cos⁡(u)\cos(u).

    So at this stage, we have dydu=cos⁡(u)=cos⁡(log⁡x)\frac{dy}{du} = \cos(u) = \cos(\log x).

  3. Differentiate the inner function with respect to xx.

    The inner function is u=log⁡xu = \log x. Its derivative is dudx=1x\frac{du}{dx} = \frac{1}{x}.

    Note

    Here log⁡x\log x means the natural logarithm (base ee), as is standard in calculus. Its derivative is 1/x1/x.

  4. Apply the Chain Rule.

    The Chain Rule says: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

    Substituting what we have: …

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