Inverse trigonometric functions(Calculus)
u=Arctan x/a ans: du/dx=a/a^2+x^2
u= Arcsin x/a, a0ans: du/dx= 1/sq. root a^2-x^2

Question

Inverse trigonometric functions(Calculus)
u=Arctan x/a      ans: du/dx=a/a^2+x^2
u= Arcsin x/a, a>0ans: du/dx= 1/sq. root a^2-x^2

anonymous · Answer

looks good to me

anonymous · Answer

we can go through it step by step if you like, as soon as the site stops having a fit

anonymous · Answer

yes pls satellite..

anonymous · Answer

i solved it but the answer is different from the the answer in the book

anonymous · Answer

first thing to know is that by the chain rule, if you want the derivative of 
$$f^{-1}(x)$$ so find it by the following :
$$\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$$

anonymous · Answer

that is because 
$$f(f^{-1}(x))=x$$ and so 
$$\frac{d}{dx}f(f^{-1}(x))=1=f'(f^{-1}(x))\times \frac{d}{dx}f^{-1}(x)$$ by the chain rule.   solving this for 
$$\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$$\]g

anonymous · Answer

sorry site if freaking out on my tonight

anonymous · Answer

am satellite can i show my solution in the 2nd equation? and can you correct it pls..bec. i can understant better in application.

anonymous · Answer

sure go ahead.

anonymous · Answer

|dw:1316625833046:dw|