what is the derivative of $\sec^2x$

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I'm using the quotient rule and so I rewrite $\sec^2x$ as $\frac{1}{\cos^2x}$

Then set $u = 1,$ $u'= 0$then $v = \cos^2x$ and $v'= -2\sin x \cos x$

I then get $\frac{2\sin x \cos x}{\cos^4x}$

but when I simplify that I don't get $2\sec^2x\tan x$ which is the correct answer? I'm not really sure what I'm doing wrong

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4 Answers

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$$\frac{2\sin(x)\cos(x)}{\cos^4(x)} = 2\cdot\frac{\sin(x)}{\cos( x)}\cdot\frac{1}{\cos^2(x)} = 2\tan(x)\sec^2(x)$$

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

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$$\frac{2\sin x \cos x}{\cos^4x}=\frac{2(\sin x)}{(\cos x)[\cos^2x]}=2\tan x \sec^2 x$$

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You simplified wrong. In $$\frac{2 \sin x \cos x}{\cos^4x}$$ start by cancelling $\cos x$ from top and bottom. Now its

$$\frac{2 \sin x }{\cos^3x}.$$

That's the same as your answer.

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