Find $dy/dx$ where $(7x+2y)^2=6x^4y^3$

Question

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.

Answer 1

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

Answer 2

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 )

Answer 3

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!