How do you differentiate #f(x)=(x/sin(7x))^2# using the chain rule?

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Answer 1 · 2016-01-17T14:05:40

#f'(x)=(2x(sin(7x)-7xcos(7x)))/sin^3(7x)#

First issue: the power of #2#.

#f'(x)=(2x)/(sin(7x))*d/dx[x/sin(7x)]#

Now, use quotient rule to find the internal derivative.

#d/dx[x/sin(7x)]=(sin(7x)d/dx[x]-xd/dx[sin(7x)])/sin^2(7x)#

Note that finding #d/dx[sin(7x)]# will require the chain rule, insofar as #d/dx[sin(u)]=cos(u)*(du)/dx#.

#d/dx[x/sin(7x)]=(sin(7x)xx1-x(cos(7x)xx7))/sin^2(7x)#

Put this back into the original expression.

#f'(x)=(2x)/sin(7x)*(sin(7x)-7xcos(7x))/sin^2(7x)#

#f'(x)=(2x(sin(7x)-7xcos(7x)))/sin^3(7x)#