# Help with using some trig identities

Need some help with the steps in converting the derivatives of the following functions.

1. derivative of $\cos(\tan(x))$ to $\frac{-\sin(\tan (x))}{\cos^2(x)}$

I can get $-\sec^2(x) \cdot (\sin(\tan(x))$ using chain rule, but then I am stuck. I guess I just need help on understanding how $\sec^2(x) = \frac{1}{\cos^2(x)}$

2. derivative of $\sin(x)\tan(x)$ to $\sin(x) + \tan(x) \cdot \sec(x)$

I can get $\cos(x) \cdot \sec^2(x)$ but then I am unsure what to do. Thanks for any help!

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## 2 Answers

For your first question, remember that $\sec$ is the reciprocal of $\cos$ by definition.

For the second, you should use the product rule to find the derivative, don't simply take the derivative of each factor. Then use the fact that $\tan{x}=\frac{\sin{x}}{\cos{x}}$ to simplify your expression and get the desired form. Try rewriting every function in terms of $\sin$ and $\cos$ to most easily simplify.

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Thanks, I see using product rule I get sinx(1/cox^2(x)) + (sinx/cosx)(cosx)  which simplifies to sinx + tanxsecx – Finzz Jan 30 '11 at 22:31
@Finzz, exactly, nice job. – yunone Jan 30 '11 at 23:04
1. $\sec(x)$ is defined to be $\frac{1}{\cos(x)}$, that is the ratio of the length of the hypotenuse and the length of the adjacent side.

2. Note that $\tan(x)\cdot \sec(x) = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)} = \sin(x) \cdot \frac{1}{\cos^2(x)}$.

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