**Toolbox:**

- Assume $ cos^{-1} \bigg( \large\frac{3cosx-2sinx}{\sqrt{13}} \bigg) = u \: and $
- $ sin^{-1} \bigg( \large\frac{5sinx+4cosx}{\sqrt{41}} \bigg) = V$
- Express $ \large\frac{3cosx-2sinx}{\sqrt{13}}$ in terms of cos and $ \large\frac{5sinx+4cosx}{\sqrt{41}}$ in terms of sin and find $ \large\frac{du}{dv}$
- $ \large\frac{du}{dv}=\large\frac{\large\frac{du}{dx}}{\large\frac{dv}{dx}}$
- $ cosAcosB-sinAsinB=cos(A+B)$
- $ sinAcosB-cosAsinB=sin(A+B)$

Let $ u = cos^{-1}\large\frac{3cosx-2sinx}{\sqrt{13}}v

$ v = sin^{-1}\large\frac{5sinx+4cosx}{\sqrt{41}}$

take $ \large\frac{3}{\sqrt{13}}=cosA\: then \: \large\frac{2}{\sqrt{13}}=sinA$

and $ \large\frac{5}{\sqrt{41}}=cosB\: then \: \large\frac{4}{\sqrt{41}}=sinB$

$ u = A+x\: and \: v=B+x$

$ u = cos^{-1}\large\frac{3}{\sqrt{13}}+x \; and \: v = cos^{-1}\large\frac{5}{\sqrt{41}}+x$

$\large \frac{du}{dx}=1\: and \:\large \frac{dv}{dx}=1$

$ \large\frac{du}{dv}=1$