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\tan^{-1}(x^2 y) = x + xy^2

Question:

{eq}\tan^{-1}(x^2 y) = x + xy^2 {/eq}

Find {eq}\frac{dy}{dx} {/eq}

Differentiation:

Explicit Differentiation :

In explicit differentiation we need to differentiate the equation firstly as it is. Then rearrange it in order to obtain y'.

After separating y', opposite one is the answer required.

Answer and Explanation:

We have,

{eq}\tan^{-1}(x^2 y) = x + xy^2 {/eq}

Now,

Differentiating both sides,

{eq}(\tan^{-1}(x^2 y))' = (x + xy^2)' \\ \displaystyle \frac{1}{1 + (x^2 y)^2} (x^2 y)' = 1 + (x'y^2 + x(y^2)') \\ \displaystyle \frac{1}{1 + (x^2 y)^2} (2x y + x^2 y') = 1 + (y^2 + x(2yy')) \\ \displaystyle y' \frac{x^2 - 2xy}{1 + (x^2 y)^2} = 1 + y^2 - \frac{1}{1 + (x^2 y)^2} (2x y) \\ \displaystyle y' = \frac{1 + (x^2 y)^2}{x^2 - 2xy}(1 + y^2 - \frac{2xy}{1 + (x^2 y)^2} ) \\ {/eq}

so,

{eq}\therefore \color{blue}{\displaystyle y' = \frac{1 + (x^2 y)^2}{x^2 - 2xy}(1 + y^2 - \frac{2xy}{1 + (x^2 y)^2} ) } {/eq}


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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