Question

differentiate w.r.t. x
sin^-1 (xsqrtx)

Answer 1

just to clarify and avoid confusion is it \(\large \frac{1}{sin(x)}\) or arcsine?

Answer 2

eitherway, you are going to be using the chain rule.

Answer 3

dats sine inverse.

Answer 4

\[\bf \frac{ dy }{ dx }\sin^{-1}(u)=\frac{ 1 }{ \sqrt{1-u^2} }\frac{ du }{ dx }\]

Answer 5

oh yes yes .. !! got it .. thanq @genius12 ... :)

Answer 6

You could derive it manually so that you don't have to memorise it. That's what I do. I always look at proofs for the derivation of a certain derivative so I don't have to memorise anything. Here is the derivation:\[\bf y =\sin^{-1}(x) \implies \sin(y) = x\]\[\bf \frac{ d }{ dx }\sin(y)=\frac{ d }{ dx }x \implies \frac{ dy }{ dx } \cos(y)=1 \implies \frac{ dy }{ dx }=\frac{ 1 }{ \cos(y) }\]\[\bf =\frac{ 1 }{ \sqrt{1-\sin^2(y)} }\]Since we declared sin(y) = x, we can re-write as:\[\bf =\frac{ 1 }{ \sqrt{1-x^2} }\]

Answer 7

@tanjeetsarkar96