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Differentiate f(x)=cos⁡(ln⁡6x)f(x)=\cos(\ln 6x)

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Q. Differentiate f(x)=cos⁡(ln⁡6x)f(x)=\cos(\ln 6x)
  1. Identify outer function: Identify the outer function and its derivative.\newlineThe outer function is the cosine function, and the derivative of cos⁡(u)\cos(u) with respect to uu is −sin⁡(u)-\sin(u).
  2. Identify inner function: Identify the inner function and its derivative.\newlineThe inner function is ln⁡(6x)\ln(6x), and the derivative of ln⁡(u)\ln(u) with respect to uu is 1/u1/u. Therefore, the derivative of ln⁡(6x)\ln(6x) with respect to xx is 1/(6x)1/(6x) multiplied by the derivative of 6x6x with respect to xx, which is 66.
  3. Apply chain rule: Apply the chain rule.\newlineThe chain rule states that the derivative of a composite function f(g(x))f(g(x)) is f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x). Here, f(u)=cos⁡(u)f(u) = \cos(u) and g(x)=ln⁡(6x)g(x) = \ln(6x), so we need to multiply the derivative of the outer function by the derivative of the inner function.
  4. Perform multiplication: Perform the multiplication to find the derivative.\newlinef′(x)=−sin⁡(ln⁡(6x))⋅(16x⋅6)f'(x) = -\sin(\ln(6x)) \cdot \left(\frac{1}{6x} \cdot 6\right)
  5. Simplify expression: Simplify the expression.\newlinef′(x)=−sin⁡(ln⁡(6x))⋅(66x)f'(x) = -\sin(\ln(6x)) \cdot \left(\frac{6}{6x}\right)\newlinef′(x)=−sin⁡(ln⁡(6x))⋅(1x)f'(x) = -\sin(\ln(6x)) \cdot \left(\frac{1}{x}\right)

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