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find dy/dx when y= (sin^-1 (x))^x + x^x

 
find dy/dx when
y=(sin^-1 (x))^x + x^x

Grade:12

1 Answers

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

y = (sin^{-1}x)^{x} + x^{x}
y = e^{log(sin^{-1}x)^{x}} + e^{log(x)^{x}}
y = e^{xlog(sin^{-1}x)} + e^{xlog(x)}
Now simply apply the chain rule
u_{1} = xlog(sin^{-1}x), u_{2} = xlog(x)
y = e^{xlog(sin^{-1}x)}.(log(sin^{-1}x)+x.\frac{1}{sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}) + e^{xlog(x)}(log(x) + x.\frac{1}{x})y = e^{xlog(sin^{-1}x)}.(log(sin^{-1}x)+\frac{x}{sin^{-1}x.\sqrt{1-x^{2}}}) + e^{xlog(x)}(log(x) + 1)
y = (sin^{-1}x)^{x}(log(sin^{-1}x)+\frac{x}{sin^{-1}x.\sqrt{1-x^{2}}}) + x^{x}.(log(x) + 1)y = (sin^{-1}x)^{x}.(log(sin^{-1}x)+\frac{x}{sin^{-1}x.\sqrt{1-x^{2}}}) + x^{x}.(log(x) + 1)y = (sin^{-1}x)^{x}.(log(sin^{-1}x)+\frac{x}{sin^{-1}x.\sqrt{1-x^{2}}}) + x^{x}.(1 + log(x))

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