Question

Hello :) Could someone help me with
what is the derivative of tan(arcsin(x)) ? :3

Answer 1

Use the chain rule
\[\Large \frac{ d }{dx } f[ g(x) ] =g'(x) * f' [g(x)]\]
think of it as derivative of the inner function (the arcsinx) multiplied by the derivative of the outer function (the tan)

You could also modify it first using trig, then differentiate the result...

Answer 2

It seems like this may come in handy in a trig conversion of the function.
\( \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} \)
Inverse sine of both sides:
\( \theta = \arcsin\left( \dfrac{\text{opposite}}{\text{hypotenuse}} \right) \)
We have \( \arcsin x\), which is equivalent to \( \arcsin \dfrac{x}{1} \).

So if we make a triangle with an angle theta, an opposite side of length x, and a hypotenuse of length 1; we can find the adjacent side by Pythagorean Thm and find \( \tan \theta \) using the trig function definition of tangent: \( \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} \)

Answer 3

Now find that
tan(arcsin x) = tan y = x/sqrt(1-x^2)

Answer 4

So just differentiate\[\Large \frac{ x }{ \sqrt{1-x^2} } \]

Answer 5

I'm confused on how "tan y = x/sqrt(1-x^2) " ? :/

Answer 6

@agent0smith ?

Answer 7

Find tan y from the triangle.

Answer 8

tan = opp/adj

Answer 9

ohhhh I see it now, Thank you! :)