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y=tan inverse(4x-4x^3/1+x^4+6x^2) then find dy/dx

y=tan inverse(4x-4x^3/1+x^4+6x^2)
then find dy/dx

Grade:12th pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

y = tan^{-1}(\frac{4x-4x^3}{1+x^4+6x^2})
u = \frac{4x-4x^3}{1+x^4+6x^2}
Use the chain rule
y' = \frac{1}{1+u^2}
y' = \frac{\frac{d}{dx}(\frac{4x-4x^3}{1+x^4+6x^2})}{1+(\frac{4x-4x^3}{1+x^4+6x^2})^2}
Take out the numerator:
{\frac{d}{dx}(\frac{4x-4x^3}{1+x^4+6x^2})} = \frac{(1+x^4+6x^2)(4-12x^2)-(4x-4x^3)(4x^3+12x)}{(1+x^4+6x^2)^2}
y ' = \frac{4(x^6-9x^4-9x^2+1)}{x^8+28x^6+6x^4+28x^2+1}


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