Calculus - derivatives

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Re: Calculus - derivatives

First your function is ambiguous. What you wrote is $$sin(x)+ \frac{3}{cos(x)}$$ but I suspect you mean $$\frac{sin(x)+ 3}{cos(x)}$$.

Second, where did get this problem? I assume from a Calculus class but then you should know
1) that the derivative of sin(x) is cos(x)
2) that the derivative of cos(x) is -sin(x)
3) the quotient rule: $$\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$$.

If you mean $$\frac{sin(x)+ 3}{cos(x)}$$ then $$f(x)= sin(x)+ 3$$ and $$g(x)= cos(x)$$.
If really do you mean $$sin(x)+ \frac{3}{cos(x)}$$ then apply the quotient rule to $$\frac{3}{cos(x)}$$ with $$f(x)= 3$$ and $$g(x)= cos(x)$$.

HallsofIvy

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