1. ## Inverse Function Derivative

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)

2. Originally Posted by keemariee

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)
$\frac{D[arcsin(u(x))]}{dx}=\frac{u'(x)}{\sqrt{1-u^2(x)}}$

I hope that helps

I will do three

D) $\frac{D[arcsin(\ln(x)-x^3)]}{dx}$....I think that is what it says

So we would get by the above formula $\frac{(ln(x)-x^3)'}{\sqrt{1-(\ln(x)-x^3)^2}}=\frac{\frac{1}{x}-3x^2}{\sqrt{1-\ln^2(x)+2\ln(x)x^3-x^6}}$

H) $\frac{D[e^{arcsin(x)}]}{dx}=(arcsin(x))'e^{arcsin(x)}=\frac{e^{arcsin(x) }}{\sqrt{1-x^2}}$

and

$\frac{D[\ln(arccos(x))]}{dx}=\frac{(arcos(x))'}{arcos(x)}=\frac{\frac{-1}{\sqrt{1-x^2}}}{arcos(x)}=\frac{-1}{\sqrt{1-x^2}arcos(x)}$

3. Originally Posted by keemariee

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)
I'll give hints, but I won't solve them:

a) Recall that $\frac{d}{dx}cos^{-1}u=-\frac{d}{dx}sin^{-1}u=-\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$

b) note that $\left(cos(x^2+x)\right)^{-1}\neq cos^{-1}(x^2+x)$

c) Recall that $\frac{d}{dx}sin^{-1}u=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$

d) note that $\left(sin(ln(x)-x^3)\right)^{-1}\neq sin^{-1}(ln(x)-x^3)$

e) Apply chain rule. It should be easy now, since I gave you the derivative of $cos^{-1}u$

f) Apply chain rule. Note that $\left(cos(x)\right)^{-1}\neq cos^{-1}(x)$

g) Apply chain rule. It should be easy, since I gave you the derivative of $sin^{-1}u$.

h) Apply chain rule. Note that $\left(sin(x)\right)^{-1}\neq sin^{-1}(x)$