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Find dy / dx in terms of x and y, if x^4 + y^2 = square root {15}.

Question:

Find {eq}\dfrac {dy}{dx} {/eq} in terms of x and y, if {eq}x^4 + y^2 = \sqrt {15} {/eq}.

Differentiation:

This is a question of explicit differentiation. So firstly we need to differentiate then by rearranging try to separate dy/dx.

Hence we will get required answer.

But in this question there is a term in square root. So we should have to remove root in order to avoid complexity.

In that case square both sides and remove root term.

Answer and Explanation:

We have,

{eq}x^4 + y^2 = \sqrt {15} {/eq}

squaring both sides,

{eq}(x^4 + y^2)^2 = 15 \\ x^8 + y^4 + 2x^4y^2 = 15 {/eq}

Now,

Differentiating with respect to x both sides,

{eq}\displaystyle 8x^7 + 4y^3 \ \frac{dy}{dx} + 8x^3y^2 + 4x^4y \ \frac{dy}{dx} = 0 \\ \displaystyle (4y^3 + 4x^4y) \ \frac{dy}{dx} = -( 8x^3y^2 + 8x^7) \\ \displaystyle \frac{dy}{dx} = - \frac{( 8x^3y^2 + 8x^7)}{(4y^3 + 4x^4y)} \\ {/eq}

so,

{eq}\therefore \color{blue}{\displaystyle \frac{dy}{dx} = - 2 \frac{( x^3y^2 + x^7)}{(y^3 + x^4y)} } {/eq}


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