Let f(x)=e^x sin(x^2). Find f'(x)

Since f(x) is a product of the two functions e^x and sin(x^2), we can use the product rule which states that if f(x)=g(x)h(x), then f'(x)=g'(x)h(x)+g(x)h'(x). Let g(x)=e^x and h(x)=sin(x^2). Since the differential of e^x is e^x, g'(x)h(x)=e^x sin(x^2), which is the first part of f'(x). For the second part of f'(x), g(x)h'(x), e^x is not differentiated, but we must use the chain rule to differentiate sin(x^2). The chain rule states that if h(x)=u(v(x)), then h'(x)=v'(x)u'(v(x)). Let u(x)=sin(x) and v(x)=x^2, differentiating x^2 using the power rule gives v'(x)=2x, and differentiating sin(x) gives u'(x)=cos(x), so u'(v(x))=cos(x^2), and h'(x)=v'(x)u'(v(x))=2xcos(x^2). This means that g(x)h'(x)=e^x 2xcos(x^2), so f'(x)=e^x sin(x^2)+e^x 2xcos(x^2).

Answered by Michael B. Maths tutor

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