# How do you find the derivative of Inverse trig function f(x) = arcsin (9x) + arccos (9x)?

##### 1 Answer
Jun 21, 2015

Here'/ the way I do this is:
- I'll let some $\text{ "theta=arcsin(9x)" }$ and some $\text{ } \alpha = \arccos \left(9 x\right)$

• So i get, $\text{ "sintheta=9x" }$ and $\text{ } \cos \alpha = 9 x$

• I differentiate both implicitly like this:
=>(costheta)(d(theta))/(dx)=9" "=>(d(theta))/(dx)=9/(costheta)=9/(sqrt(1-sin^2theta))=9/(sqrt(1-(9x)^2)

-- Next, I differentiate $\cos \alpha = 9 x$
$\implies \left(- \sin \alpha\right) \cdot \frac{d \left(\alpha\right)}{\mathrm{dx}} = 9 \text{ } \implies \frac{d \left(\alpha\right)}{\mathrm{dx}} = - \frac{9}{\sin \left(\alpha\right)} = - \frac{9}{\sqrt{1 - \cos \alpha}} = - \frac{9}{\sqrt{1 - {\left(9 x\right)}^{2}}}$

• Overall, $\text{ } f \left(x\right) = \theta + \alpha$

• So, ${f}^{' '} \left(x\right) = \frac{d \left(\theta\right)}{\mathrm{dx}} + \frac{d \left(\alpha\right)}{\mathrm{dx}} = \frac{9}{\sqrt{1 - {\left(9 x\right)}^{2}}} - \frac{9}{\sqrt{1 - {\left(9 x\right)}^{2}}} = 0$