Solve for the one with 1,2,3 and 4

Mathematics

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Here's the solution for the question with options 1, 2, 3, and 4 (Question 63).

63. If y=sin⁡(x2)y = \sin(x^2), find dydx\frac{dy}{dx}.

  1. Identify the functions for the chain rule.

    Let u=x2u = x^2. Then y=sin⁡(u)y = \sin(u).

  2. Find the derivatives of yy with respect to uu and uu with respect to xx.

    dydu=ddu(sin⁡u)=cos⁡u\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u dudx=ddx(x2)=2x\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x
  3. Apply the chain rule, dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

    dydx=(cos⁡u)⋅(2x)\frac{dy}{dx} = (\cos u) \cdot (2x)
  4. Substitute u=x2u = x^2 back into the expression.

    dydx=2xcos⁡(x2)\frac{dy}{dx} = 2x \cos(x^2)

    The correct option is 1. 2xcos⁡(x2)2x \cos(x^2).

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