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Composite Function

The value of ...

Question

The value of dy/dx is, when y=f(1/x) and f'(x)=sin(x²)

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Solution

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y = f (\frac{2 x - 1}{x^{2} + 1})

f^{'} (x) = sin x

\frac{d y}{d x}

A = \frac{d y}{d x}

x^{2} + y^{2} = 4

(\sqrt{2}, \sqrt{2})

B = \frac{d y}{d x}

sin y + sin x = sin x - sin y

(π, π)

C = \frac{d y}{d x}

2 e^{x y} + e^{x} e^{y} - e^{x} - e^{y} = e^{x y + 1}

(1, 1)

(A + B + C)

\frac{d y}{d x}

y = e^{\sqrt{x} + sin x}

y = {tan}^{- 1} \sqrt{\frac{1 - sin x}{1 + sin x}}

\frac{d y}{d x}

x = \frac{π}{6}

If $y = t a n^{- 1} \sqrt{\frac{1 - s i n x}{1 + s i n x}}$ then the value of $\frac{d y}{d x}$ at $x = \frac{π}{6}$ is

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