Find derivative of 5^(arcsine(x))

  • Thread starter Thread starter oceanflavored
  • Start date Start date
  • Tags Tags
    Derivative
oceanflavored
Messages
44
Reaction score
0

Homework Statement


how do you take the derivative of y = 5^(inverse sine of x)?

Homework Equations


the formula to take the derivative of arcsin x = 1 / square root of 1 - X2

The Attempt at a Solution


i tried to use the chain rule so that i did...

5 ^ (inverse sin of x) times (the derivative formula)

but i don't think it's right!

PLEASE HELP!

thanks bunches :)
 
Physics news on Phys.org
You have to use implicit differentiation. Firstly you have to "bring the arcsin x down" by taking the natural logarithm of both sides. Then it becomes

ln|y|=sin^{-1}x(ln5)

Then simply differentiate implicitly.
 
That is certainly one way to do it and probably simplest but it is not necessary to use "logarithmic differentiation", just the chain rule.

The derivative of ax, with respect to x, for any positive number a, is ln(a)a^{x}. The derivative of
a^{sin^{-1}(x)}[/itex] <br /> is, of course,<br /> ln(a) a^{sin^{-1}(x)} \frac{d}{dx}sin^{-1}(x)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top