Question

hi can someone pls. explain to me about Inverse trigonometric functions(Differential Calculus).
pls. solve this problem then explain pls.
y=square root of Arcsinx

y=xArcsin2x

Answer 1

now clear what the problem is so "solve" are you looking for the derivatives of these?

Answer 2

yes

Answer 3

that is wrong.
\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 4

in this case
\[f(x)=\arcsin(x)\] so
\[f'(x)=\frac{1}{\sqrt{1-x^2}}\]

Answer 5

so the first step is to write
\[\frac{d}{dx}\sqrt{\arcsin(x)}=\frac{\frac{1}{\sqrt{1-x^2}}}{2\sqrt{\arcsin(x)}}\]

Answer 6

then some algebra maybe to clean it up

Answer 7

for
\[y=x\arcsin(2x)\] you need the product rule and the chain rule

Answer 8

product rule use
\[(fg)'=f'g+g'f\]
here
\[f(x)=x,f'(x)=1,g(x)=\arcsin(x),g'(x)=\frac{2}{\sqrt{1-4x^2}}\]

Answer 9

put it together in the product rule to get the answer

Answer 10

tnx satellite