If f(2)=-3, f'(2)=3/4 and g(x)=f'(x), what is the equation of the tangent line in g(x) at x=-3?

If f(2)=-3, f'(2)=3/4 and g(x)=f'(x), what is the equation of the tangent line in g(x) at x=-3?

Just need the first derivative of this function, struggling with there to begin

This question is from isc class 12 book

A) y+2=4/3(x-3) B) y-2=4/3(x+3) C) y-2=-3/4(x+3) D) y-2=-4/3(x+3) E) y+2=-3/4(x-3)

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Please help me differentiates this.

Pretty straight forward, but i am having trouble. You help would be much appreciated

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