It's not cos(5x-3). You've got to use the chain rule. Remember? Derivative of f(g(x)) is not f'(g(x)).
f(x) = sin(x); g(x)=5x-3.
Quote:
Theorem. Let F be the composition of two differentiable functions f and g; F(x) = f(g(x)). Then F is differentiable and
F'(x) = f '(g(x)) g '(x)
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Therefore, the derivative of sin(5x-3) is:
[ f'(x)=cos(x) ; g'(x)=5 ]
5*cos(5x-3).
You're welcome
