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