Question

tan inverse 2x/1+15x^2

Answer 1

What has to be done here?

Answer 2

\[\tan^{-1} \frac{2x}{1+15x^2}\]

Answer 3

differentiate

Answer 4

Ok pls wait

Answer 5

Do you know the differentiation of \[\tan^{-1}x\]

Answer 6

yes

Answer 7

It is \[\tan^{-1}x = \frac {1}{1+x^2}\]
so using it we can

Answer 8

can u solve it if u dont mind

Answer 9

Therefore
let \[y= \tan^{-1} \frac{2x}{1+15x^2} \rightarrow y' = \frac{d}{dx}\tan^{-1} \frac{2x}{1+15x^2}\]
\[y' = \frac{1}{1+(\frac{2x}{1+15x^2})^2} \frac{d}{dx}(\frac{2x}{1+15x^2})\]
\[y' = \frac{1}{1+\frac{4x^2}{(1+15x^2)^2}} \frac{d}{dx}(\frac{2x}{1+15x^2})\]\[y' = \frac{1}{\frac{(1+15x^2)^2+4x^2}{(1+15x^2)^2}} \frac{d}{dx}(\frac{2x}{1+15x^2})\]
\[y' = \frac{(1+15x^2)^2}{(1+15x^2)^2+4x^2} (\frac{(1+15x^2) \frac{d}{dx}2x-2x \frac{d}{dx}(1+15x^2)}{(1+15x^2)^2})\]
\[y' = \frac{(1+15x^2)^2}{(1+15x^2)^2+4x^2} (\frac{(1+15x^2) 2-2x (30x)}{(1+15x^2)^2})\]
\[y' = (\frac{(1+15x^2) 2-2x (30x)}{(1+15x^2)^2+4x^2})\]
\[y' = (\frac{2+30x^2-60x^2}{(1+15x^2)^2+4x^2})\]
\[y' = \frac{2-30x^2}{(1+15x^2)^2+4x^2}= \frac{2-30x^2}{1+225x^4+30x^2+4x^2}\]
\[y' = \frac{2-30x^2}{1+225x^4+34x^2}\]

Answer 10

@shalinisengupta

Answer 11

thanks :)

Answer 12

Medal please

Answer 13

i m new so dont know how to give medal

Answer 14

@shalinisengupta