Differentiate Tan^-1y=x^3?

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Differentiate Tan^-1y=x^3?

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Answer 1 · 2017-09-02T15:35:32

$\frac{d y}{d x} = 3 x^{2} {sec}^{2} (x^{3})$ $dy/dx=3x^2sec^2(x^3)$

Explanation:

${tan}^{- 1} y = x^{3}$ $tan^-1y=x^3$

Method 1 - No simplification

Differentiate as written. Recall that $\frac{d}{d x} {tan}^{- 1} x = \frac{1}{1 + x^{2}}$ $d/dxtan^-1x=1/(1+x^2)$ . Here, differentiating ${tan}^{- 1} y$ $tan^-1y$ with respect to $x$ $x$ will cause the chain rule to be in effect.

$\frac{1}{1 + y^{2}} \cdot \frac{d y}{d x} = 3 x^{2}$ $1/(1+y^2)*dy/dx=3x^2$

Solving for the derivative:

$\frac{d y}{d x} = 3 x^{2} (1 + y^{2})$ $dy/dx=3x^2(1+y^2)$

Let's write this all in terms of $x$ $x$ . To do this, we have to solve for $y$ $y$ . From ${tan}^{- 1} y = x^{3}$ $tan^-1y=x^3$ , note that $y = tan (x^{3})$ $y=tan(x^3)$ .

$\frac{d y}{d x} = 3 x^{2} (1 + {tan}^{2} (x^{3}))$ $dy/dx=3x^2(1+tan^2(x^3))$

Note that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ :

$\frac{d y}{d x} = 3 x^{2} {sec}^{2} (x^{3})$ $dy/dx=3x^2sec^2(x^3)$

Method 2 - Simplification

Recall that $y = tan (x^{3})$ $y=tan(x^3)$ . We can then differentiate this directly, which is easier. It helps to know that $\frac{d}{d x} tan x = {sec}^{2} x$ $d/dxtanx=sec^2x$ . We will use that here and also use the chain rule.

$y = tan (x^{3})$ $y=tan(x^3)$

$\frac{d y}{d x} = {sec}^{2} (x^{3}) \cdot \frac{d}{d x} x^{3}$ $dy/dx=sec^2(x^3)*d/dxx^3$

$\frac{d y}{d x} = 3 x^{2} {sec}^{2} (x^{3})$ $dy/dx=3x^2sec^2(x^3)$

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Differentiate Tan^-1y=x^3?

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