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Worked Examples · Example 21

Q.Find the derivative of the function given by f(x)=sin⁡(x2)f(x) = \sin(x^2).

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The derivative of sin⁡(x2)\sin(x^2) is found using the Chain Rule: differentiate the outer sine function, then multiply by the derivative of the inner x2x^2. The result is f′(x)=2xcos⁡(x2)f'(x) = 2x \cos(x^2).

The Chain Rule is the natural tool here because f(x)=sin⁡(x2)f(x) = \sin(x^2) is a composition of two functions. You have an outer function, sin⁡(u)\sin(u), and an inner function, u=x2u = x^2. The derivative of a composition is: derivative of the outer function (evaluated at the inner function) times the derivative of the inner function. In Leibniz notation, dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

Why does this work? Because the rate of change of ff with respect to xx depends on how fast sin⁡(u)\sin(u) changes with uu, and how fast uu itself changes with xx. You multiply these rates.

Let’s apply it step by step.

  1. Identify the outer and inner functions.

    Outer: sin⁡(u)\sin(u), where uu is a placeholder.

    Inner: u=x2u = x^2.

  2. Differentiate the outer function with respect to its argument.

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

    So ddusin⁡(u)=cos⁡(u)\frac{d}{du} \sin(u) = \cos(u).

  3. Differentiate the inner function with respect to xx.

    The derivative of x2x^2 is 2x2x.

    So dudx=2x\frac{du}{dx} = 2x.

  4. Apply the Chain Rule.

    Multiply the two derivatives, but remember: the outer derivative must be evaluated at the original inner function u=x2u = x^2, not at uu alone. …

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