# Thread: Differentiation of Inverse Trig Function

1. ## Differentiation of Inverse Trig Function

Differention with repeat to x
cot^-1(x)

i stuck...

2. Originally Posted by sanikui
Differention with repeat to x
cot^-1(x)

i stuck...
Use integration by parts: $\displaystyle \int u \, dv = uv - \int v \, du$.

$\displaystyle u = \cot^{-1} x \Rightarrow du = \, ....$

$\displaystyle dv = dx \Rightarrow v = \, ....$

3. Originally Posted by mr fantastic
Use integration by parts: $\displaystyle \int u \, dv = uv - \int v \, du$.

$\displaystyle u = \cot^{-1} x \Rightarrow du = \, ....$

$\displaystyle dv = dx \Rightarrow v = \, ....$
My mistake. I misread differentiate as integrate.

$\displaystyle y = \cot^{-1} x \Rightarrow \cot y = x$

Now differentiate both sides wrt x (use the chain rule on the left hand side) and then make dy/dx the subject.

4. ok i try.

5. Originally Posted by mr fantastic
My mistake. I misread differentiate as integrate.

$\displaystyle y = \cot^{-1} x \Rightarrow \cot y = x$

Now differentiate both sides wrt x (use the chain rule on the left hand side) and then make dy/dx the subject.
Spoilers in white:

Using the chain rule: $\displaystyle {\color{white} \frac{d \, \cot y}{dx} = \frac{d \, \cot y}{dy} \cdot \frac{dy}{dx} = -\frac{1}{\sin^2 y} \frac{dy}{dx}.}$

Note that: $\displaystyle {\color{white} \cot y = x \Rightarrow \sin^2 y = \frac{1}{x^2 + 1}.}$

6. Finaly noe how to do this question. V^^
Let x=coty

x=1 over tany
tany= 1 over x
y= tan(1 over x)

From there i can do liaw....thx alot ^^