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Find $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}$ where

$$(7x+2y)^2=6x^4y^3$$

This is on my homework but book has different examples so I don't know what side to start on.

Paul
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sunsetsandbeachvibes
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    What have you tried so far? What kind of things do you think you should be trying to do? The more context you provide in your question, the more we'll be able to gauge your level and progress. – Semiclassical Jul 22 '14 at 23:54
  • "Use implicit differentiation" should be enough to start (you should use it on both sides). – hardmath Jul 23 '14 at 00:56

3 Answers3

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First step: that $\frac{d}{dx}$ of both sides (this means we implicitly differentiate with respect to $x$ on both sides, and we say that $$\frac{d}{dx}f(y) = \frac{d}{dy}*\frac{dy}{dx}f(y) = \frac{dy}{dx}*f'(y)$$ Moving on to the question: $$\frac{d}{dx}(7x+2y)^2 = \frac{d}{dx}(6x^4y^3)$$ $$2(7x+2y)*\frac{d}{dx}(7x+2y) = 24x^3y^3+\frac{d}{dx}(y^3)*(6x^4)$$ $$2(7x+2y)*(7 + \frac{d}{dx}(2y)) = 24x^3y^3 + \frac{dy}{dx}*18x^4y^2$$ $$2(7x+2y)(7+\frac{dy}{dx}*2) = 24x^3y^3 + \frac{dy}{dx}*18x^4y^2$$ Then it's just a simple matter of solving for $\frac{dy}{dx}$. Hope this helped!

Vishwa Iyer
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To get a better grip on the situation, write $y = f(x)$. Then it should be easier to see how the chain rule comes into play. Remember to use the product rule on the right-hand side. Start with: $$(7x + 2f(x))^2 = 6x^4(f(x))^3$$

Ivo Terek
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By the chain rule and implicit differentiation. $$\frac{\mathrm{d}}{\mathrm{d}x}\text{LHS}=2(7x+2y)\cdot\frac{\mathrm{d}}{\mathrm{d}x}(7x+2y)$$ By the product rule on the right side. $$98x+28x\frac{\mathrm{d}y}{\mathrm{d}x}+28y+8y\frac{\mathrm{d}y}{\mathrm{d}x}=18x^4y^2\frac{\mathrm{d}y}{\mathrm{d}x}+24x^3y^3$$ Solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$. $$98x+28x\frac{\mathrm{d}y}{\mathrm{d}x}+28y+8y\frac{\mathrm{d}y}{\mathrm{d}x}=18x^4y^2\frac{\mathrm{d}y}{\mathrm{d}x}+24x^3y^3$$ $$\therefore\quad\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{12x^3y^3-49x-14y}{14x+4y-9x^4y^2}$$

(Or you could use wolframalpha http://bit.ly/1yYTQk5 )

Jam
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