# Differentiate $\cos^{-1} \bigg[\large\frac{3 \cos x-2 \sin x}{\sqrt {13}}\bigg] w.r.t \; \sin^{-1} \bigg[\large\frac{5\sin x +4 \cos x}{\sqrt{41}}\bigg]$

• 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 $\large\frac{du}{dv}$
• $\large\frac{du}{dv}=\large\frac{\large\frac{du}{dx}}{\large\frac{dv}{dx}}$
• $cosAcosB-sinAsinB=cos(A+B)$
• $sinAcosB-cosAsinB=sin(A+B)$