Given that $tan y = x^{2}$ $tany= x^2$ , what is the value of $\frac{d y}{d x}$ $dy/dx$ ?

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Given that $tan y = x^{2}$ $tany= x^2$ , what is the value of $\frac{d y}{d x}$ $dy/dx$ ?

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Answer 1 · 2016-12-07T03:28:26

$\frac{d y}{d x} = \frac{2 x}{{sec}^{2} y}$ $dy/dx = (2x)/sec^2y$

Explanation:

We write $tan y$ $tany$ as $\frac{sin y}{cos y}$ $siny/cosy$ and differentiate using the quotient rule (with respect to $x$ $x$ ) .

$\frac{sin y}{cos y} = x^{2}$ $siny/cosy = x^2$

$\frac{cos y (cos y) (\frac{d y}{d x}) - (- sin y \times sin y) \frac{d y}{d x}}{{(cos y)}^{2}} = 2 x$ $(cosy(cosy)(dy/dx) - (-siny xx siny)dy/dx)/(cosy)^2 = 2x$

$\frac{{cos}^{2} y (\frac{d y}{d x}) + {sin}^{2} y (\frac{d y}{d x})}{{cos}^{2} y} = 2 x$ $(cos^2y(dy/dx) + sin^2y(dy/dx))/cos^2y = 2x$

We use the identity ${sin}^{2} β + {cos}^{2} β = 1$ $sin^2beta + cos^2beta = 1$ to simplify further...

$\frac{\frac{d y}{d x}}{{cos}^{2} y} = 2 x$ $(dy/dx)/(cos^2y) = 2x$

Use the identity $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ ...

$\frac{d y}{d x} {sec}^{2} y = 2 x$ $dy/dxsec^2y = 2x$

$\frac{d y}{d x} = \frac{2 x}{{sec}^{2} y}$ $dy/dx = (2x)/sec^2y$

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Given that $tan y = x^{2}$ $tany= x^2$ , what is the value of $\frac{d y}{d x}$ $dy/dx$ ?

1 Answer

Explanation:

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Given that tany= x^2tany=x2tany= x^2, what is the value of dy/dxdydxdy/dx?

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Given that $tan y = x^{2}$ $tany= x^2$ , what is the value of $\frac{d y}{d x}$ $dy/dx$ ?