# Multivariable calculus - implicit differentiation

• Mar 6th 2010, 06:46 PM
nyasha
Multivariable calculus - implicit differentiation
Guys l just want to know if l am on the right path in this question. The question and attempt to the solution is on the pdf file attached.
• Mar 6th 2010, 08:05 PM
Prove It
Just use the chain rule.

You should know that $\frac{du}{dx} = \frac{du}{dt}\,\frac{dt}{dx}$.

Since

$x^3t + x^2t - 4 = 0$

$\frac{d}{dx}(x^3t + x^2t - 4) = \frac{d}{dx}(0)$

$\frac{d}{dx}(x^3t) + \frac{d}{dx}(x^2t) + \frac{d}{dx}(-4) = 0$

$x^3\,\frac{dt}{dx} + 3x^2t + x^2\,\frac{dt}{dx} + 2xt + 0 = 0$

$(x^3 + x^2)\,\frac{dt}{dx} + 3x^2t + 2xt = 0$

$x^2(x + 1)\,\frac{dt}{dx} + xt(3x + 2) = 0$

$x^2(x + 1)\,\frac{dt}{dx} = -xt(3x + 2)$

$\frac{dt}{dx} = -\frac{xt(3x + 2)}{x^2(x + 1)}$

$\frac{dt}{dx} = -\frac{t(3x + 2)}{x(x + 1)}$.

You also know that

$u = e^{5 + t} + \cos{t}$

so $\frac{du}{dt} = e^{5 + t} - \sin{t}$.

Therefore

$\frac{du}{dx} = \frac{du}{dt}\,\frac{dt}{dx}$

$= -\frac{t\left(e^{5 + t} - \sin{t}\right)(3x + 2)}{x(x + 1)}$.
• Mar 7th 2010, 12:32 AM
nyasha
Quote:

Originally Posted by Prove It
Just use the chain rule.

You should know that $\frac{du}{dx} = \frac{du}{dt}\,\frac{dt}{dx}$.

Since

$x^3t + x^2t - 4 = 0$

$\frac{d}{dx}(x^3t + x^2t - 4) = \frac{d}{dx}(0)$

$\frac{d}{dx}(x^3t) + \frac{d}{dx}(x^2t) + \frac{d}{dx}(-4) = 0$

$x^3\,\frac{dt}{dx} + 3x^2t + x^2\,\frac{dt}{dx} + 2xt + 0 = 0$

$(x^3 + x^2)\,\frac{dt}{dx} + 3x^2t + 2xt = 0$

$x^2(x + 1)\,\frac{dt}{dx} + xt(3x + 2) = 0$

$x^2(x + 1)\,\frac{dt}{dx} = -xt(3x + 2)$

$\frac{dt}{dx} = -\frac{xt(3x + 2)}{x^2(x + 1)}$

$\frac{dt}{dx} = -\frac{t(3x + 2)}{x(x + 1)}$.

You also know that

$u = e^{5 + t} + \cos{t}$

so $\frac{du}{dt} = e^{5 + t} - \sin{t}$.

Therefore

$\frac{du}{dx} = \frac{du}{dt}\,\frac{dt}{dx}$

$= -\frac{t\left(e^{5 + t} - \sin{t}\right)(3x + 2)}{x(x + 1)}$.

Was the method l used correct ?
• Mar 7th 2010, 04:00 AM
HallsofIvy
Yes. Once you complete those determinants, you should get the same answer.