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# inverse trig functions derivative

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f(x) = tan^-1(2x+1)

Guest Dec 6, 2014

#4
+20025
+10

f(x) = tan^-1(2x+1)      f(x)'  ?

$$\tan[\ {\tan^{-1}(2x+1)} \ ] = 2x+1 \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\ (1+\tan^2(\ {\tan^{-1}(2x+1)} \ ) ) \times \left[\ {\tan^{-1}(2x+1)} \ \right]' = 2 \\\\ \left[ 1+(2x+1)^2} \right] \times \left[{\tan^{-1}(2x+1)} \ \right]' = 2 \\\\ \boxed{ \left[\ {\tan^{-1}(2x+1)} \ \right]' = \frac{2} {1+(2x+1)^2} } }$$

heureka  Dec 8, 2014
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+27057
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.

Alan  Dec 6, 2014
#2
+10

Thanks for your answer. I really would love some explanation and if you can show me how to solve this problem by using u-substituiton I would appreciate.

Thanks again. :)

Guest Dec 6, 2014
#3
+93683
+10

$$\\y=tan^{-1}(2x+1)\\\\ Let\;\; u=2x+1\qquad \frac{du}{dx}=2\\\\ y=tan^{-1}\;u\\\\ u=tan\;y\\\\ \frac{du}{dy}=sec^2y\\\\ \frac{dy}{du}=\frac{1}{sec^2y}\\\\$$

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$$\\y=tan^{-1}(2x+1)\\\\ Let\;u=2x+1\qquad \frac{du}{dx}=2\\\\ y=tan^{-1}u\\\\ u=tan\;y\\\\ \frac{du}{dy}=sec^2y\\\\ \frac{dy}{du}=\frac{1}{sec^2y}\\\\ \begin{array}{rll} \frac{dy}{dx}&=&\frac{dy}{du}\times\frac{du}{dx}\\\\ &=&\frac{1}{sec^2y}\times 2\\\\ &=&\frac{2}{sec^2y}\\\\ &=&refer to explanation below to help get to next step\\\\ &=&\frac{2}{1+u^2}\\\\ &=&\frac{2}{1+(2x+1)^2}\\\\ &=&\frac{2}{4x^2+4x+2}\\\\ &=&\frac{1}{2x^2+2x+1}\\\\ \end{array}$$

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tan y = u/1

Melody  Dec 7, 2014
#4
+20025
+10
$$\tan[\ {\tan^{-1}(2x+1)} \ ] = 2x+1 \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\ (1+\tan^2(\ {\tan^{-1}(2x+1)} \ ) ) \times \left[\ {\tan^{-1}(2x+1)} \ \right]' = 2 \\\\ \left[ 1+(2x+1)^2} \right] \times \left[{\tan^{-1}(2x+1)} \ \right]' = 2 \\\\ \boxed{ \left[\ {\tan^{-1}(2x+1)} \ \right]' = \frac{2} {1+(2x+1)^2} } }$$