inverse functions: d/dx cos^-1(x^2+x), d/dx (cos(x^2+x))^-1,

[kcbrat]

i'm stuck on these..
a) d/dx cos^-1(x^2+x)
b) d/dx (cos(x^2+x))^-1
c) d/dx sin^-1(ln(x)-x^3)
d) d/dx sin(ln(x)-x^3)^-1
e) d/dx ln(cos^-1(x))
f) d/dx ln((cos x)^-1)
g) d/dx exp(sin^-1(x))
h) d/dx exp((sin x)^-1)

 

[royhaas]

Re: inverse functions

Use the chain rule.

 

[galactus]

Re: inverse functions

As Roy said, the chain rule. I will go ahead and do the first so you get the gist.

The chain rule is more or less just the derivative of the outside times the derivative of the inside.

So, you have:

\(\displaystyle cos^{-1}(x^{2}+x)\)

The derivative of arccos is \(\displaystyle \frac{-1}{\sqrt{1-x^{2}}}\)

Sub in \(\displaystyle x^{2}+x\) for the x in the derivative of arccos:

'\(\displaystyle \frac{-1}{\sqrt{1-(x^{2}+x)^{2}}}\)

The derivative of \(\displaystyle x^{2}+x\) is \(\displaystyle 2x+1\)

So, putting it together:

\(\displaystyle \frac{-(2x+1)}{\sqrt{-x^{4}-2x^{3}-x^{2}+1}}\)

Or you can leave it as: \(\displaystyle \frac{-(2x+1)}{\sqrt{1-(x^{2}+x)^{2}}}\)