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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]$

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  • 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$

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 25, 2013 by thanvigandhi_1
 

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