derivative of f(x)=tan^2x^2

**jmanna98** · August 10th 2011, 04:47 PM

Can someone please show me the steps to this problem? Thanks alot

**pickslides** · August 10th 2011, 04:52 PM

Use the chain rule twice here.

Say $y = \tan ^2x^2$

Find $\frac{dy}{dx}$ where $u = \tan x^2 \implies y = u^2$

what do you get?

**Plato** · August 10th 2011, 05:05 PM

Originally Posted by jmanna98

Can someone please show me the steps to this problem? Thanks alot

The derivative of $f^2(x^2)$ is $2[f(x^2)]f'(x^2)(2x).$

**jmanna98** · August 10th 2011, 07:02 PM

f(x)=tan^2x^2
f'(x)=(tanx^2)^2
f'(x)=2(tanx^2)[sec(x^2)(2x)]

Where am I making errors? I am unsure if I am understanding the chain rule properly for trig functions.

My understanding is that you take the derivative of the outside function multiplied by the inside function and multiplied by the derivative of the inside function.

**Bushy** · August 10th 2011, 07:07 PM

Originally Posted by jmanna98

f'(x)=2(tanx^2)[sec(x^2)(2x)]

That looks correct to me.

**TKHunny** · August 10th 2011, 07:08 PM

You're close. $\frac{d}{dx}\tan(x) = \sec^{2}(x) \ne \sec(x)$

**jmanna98** · August 10th 2011, 07:58 PM

oh...thats what I meant. My bad. Thanks for the confirmation!

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