Chain rule using trig functions

**tmas** · May 27th 2012, 11:05 AM

Find the derivative: (do not simplify)
f(x) = sec(sqrt(1+(2x/tanx)))

Is this the correct answer?
tan (sqrt(1+(2x/tanx))) sec (sqrt(1+(2x/tanx)))1/2(1+(2x/tanx)))^-1/2 ((2tanx-2xsec^x)/(tan^2x))

sorry I don't know how to make this look any better.

**Plato** · May 27th 2012, 11:28 AM

Originally Posted by tmas

Find the derivative: (do not simplify)
f(x) = sec(sqrt(1+(2x/tanx)))
Is this the correct answer?
tan (sqrt(1+(2x/tanx))) sec (sqrt(1+(2x/tanx)))1/2(1+(2x/tanx)))^-1/2 ((2tanx-2xsec^x)/(tan^2x))

sorry I don't know how to make this look any better.

Almost, The last factor is $\frac{2\tan(x)-2x\sec^2(x)}{\tan^2(x)}$

**tmas** · May 27th 2012, 11:34 AM

Why is it x^2 in the second part of the subtraction.. before taking the derivative isn't the numerator:

(2x)'(tanx) - 2x(tanx)' ? I do not quite understand...

**Plato** · May 27th 2012, 11:40 AM

That was a typo.

**Soroban** · May 27th 2012, 11:42 AM

Hello, tmas!

$\text{Find the derivative (do not simplify): }\:f(x) \:=\: \sec\!\left(\!\sqrt{1+\frac{2x}{\tan x}}\right)$

$\text{Is this the correct answer?}$

. $f'(x) \;=\;\tan\!\left(\!\sqrt{1+\frac{2x}{\tan x}}}\right)\sec\!\left(\!\sqrt{1+\frac{2x}{\tan x}}\right) \cdot\frac{1}{2}\cdot \left(1+\frac{2x}{\tan x}\right)^{-\frac{1}{2}}\!\left(\frac{2\tan x-2x\sec^2x}{\tan^2x}\right)$

Correct! . . . Good work!

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