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y= cos^-1(2x + 3 sqrt(1-x^2) / sqrt(13)) find dy/dx

y=cos^-1(2x + 3 sqrt(1-x^2) / sqrt(13)) find dy/dx

Grade:12

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:

Hello Student,
Please find answer to your question

y = cos^{-1}(\frac{2x+3\sqrt{1-x^{2}}}{\sqrt{13}})
Apply the chain rule
\frac{dy}{dx} = \frac{d(cos^{-1}u)}{dx}.\frac{du}{dx}
u = \frac{2x+3\sqrt{1-x^{2}}}{\sqrt{13}}
u^{2} = \frac{4x^2+9(1-x^{2})+12x\sqrt{1-x^{2}}}{13}
u^{2} = \frac{9-5x^{2}+12x\sqrt{1-x^{2}}}{13}
1-u^{2} = 1-\frac{9-5x^{2}+12x\sqrt{1-x^{2}}}{13}
1-u^{2} = \frac{4+5x^{2}-12x\sqrt{1-x^{2}}}{13}

\frac{du}{dx} = \frac{2+3.\frac{-x}{\sqrt{1-x^{2}}}}{\sqrt{13}}
\frac{du}{dx} = \frac{2\sqrt{1-x^{2}}-3x}{\sqrt{13(1-x^{2})}}
\frac{dy}{dx} = \frac{d(cos^{-1}u)}{dx}.\frac{du}{dx}
\frac{dy}{dx} = \frac{-1}{\sqrt{1-u^{2}}}.\frac{du}{dx}
\frac{dy}{dx} = \frac{-\sqrt{13}}{\sqrt{4+5x^{2}-12x\sqrt{1-x^{2}}}}.\frac{du}{dx}
\frac{dy}{dx} = \frac{-\sqrt{13}}{\sqrt{4+5x^{2}-12x\sqrt{1-x^{2}}}}.\frac{2\sqrt{1-x^{2}}-3x}{\sqrt{13(1-x^{2})}}
\frac{dy}{dx} = \frac{3x-2\sqrt{1-x^{2}}}{\sqrt{4+5x^{2}-12x\sqrt{1-x^{2}}}}.\frac{1}{\sqrt{1-x^{2}}}
\frac{dy}{dx} = \frac{3x-2\sqrt{1-x^{2}}}{\sqrt{(3x-2\sqrt{1-x^{2}})^{2}}}.\frac{1}{\sqrt{1-x^{2}}}\frac{dy}{dx} = \frac{3x-2\sqrt{1-x^{2}}}{|(3x-2\sqrt{1-x^{2}})|}.\frac{1}{\sqrt{1-x^{2}}}
\frac{dy}{dx} = \frac{\pm 1}{\sqrt{1-x^{2}}}

Zaid
11 Points
3 years ago
Above answer is very lengthy ..
I can tell u the easiest method
y=cos^-1(2/sqrt13 (x) - 13/sqrt13  sqrt(1-x^2)
Let cosA=2/sqrt13 , cosB=x
      SinA =3/sqrt13 , sinB= sqrt(1-x^2)
Therefore,
y= cos^-1(cosAcosB-sinAsinB)
   = cos^-1[cos(A+B)]
   = A+B
   = sin^-1(3/sqrt13) + cos^-1(x)
   = 0 + ( - 1/sqrt(1-x^2) )
   = -1/sqrt(1-x^2)

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