Chain Rules - Partial Derivatives

**Belowzero78** · April 13th 2010, 02:30 PM

Let $z =f(x,y) = (8+25x)y^{tan(x)})$ , where: $x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t)$ , and $y = (2 - t)e^{t} - 12(t-1)$ . Use the chain rule to find $\frac{dz}{dt}$ at $t = 1$ .

$\frac{dz}{dt} = ?$

The Chain rule formula I used is the following: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$

And, when $t = 1$ , $x = \frac{\pi}{4}$
When $t = 1$ , $y = e$

$\frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)}$

$\frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1}$ . This is one correct?

$\frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t}$

$\frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}}$

**dwsmith** · April 13th 2010, 05:36 PM

Originally Posted by Belowzero78

Let $z =f(x,y) = (8+25x)y^{tan(x)})$ , where: $x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t)$ , and $y = (2 - t)e^{t} - 12(t-1)$ . Use the chain rule to find $\frac{dz}{dt}$ at $t = 1$ .

$\frac{dz}{dt} = ?$

The Chain rule formula I used is the following: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
In your first partial, you should an x wrt t not y wrt t but you did the correct derivatives in your work.

And, when $t = 1$ , $x = \frac{\pi}{4}$
When $t = 1$ , $y = e$

$\frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)}$

$\frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1}$ . This is one correct?

$\frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t}$

$\frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}}$

Looks fine besides the typo; however, I didn't check your derivatives. I am little confused on what you are asking. You want to know what $\frac{dz}{dt}$ is but you have done everything as long as your math is correct.

**Belowzero78** · April 13th 2010, 06:32 PM

Could someone please show me what the partial is for: $\frac{\partial z}{\partial x}$ ?
I'm really confused about the y^tan(x) part when taking the derivative of x.

Would it be: $\frac{\partial z}{\partial x} = (8 + 25x)y^{\tan(x)}\ln(y)\sec^{2}(x) + 25y^{\tan(x)}$ ?

**dwsmith** · April 13th 2010, 06:50 PM

For example, $y^{tan(x)}$ can be solved like this:

$y=u$ and $tanx=v$

$\big[\frac{u'*v}{u}+ln(u)*v'\big]*u^v$

Therefore, $25y^{tan(x)}=25\frac{y^{tan(x)}*ln(y)}{cos^2(x)}$

If this example doesn't help you with, then just ask.

**Belowzero78** · April 13th 2010, 06:52 PM

Ok, after this how do i evaluate dz/dt? Like what stuff do i plug in?

**dwsmith** · April 13th 2010, 06:54 PM

Add your solutions $\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}<br />$ of the partial derivatives you obtain.

Thread: Chain Rules - Partial Derivatives

LinkBack

Thread Tools

Display

Chain Rules - Partial Derivatives

Similar Math Help Forum Discussions

Chain rule for partial derivatives

Using quotient and chain rules for derivatives

Partial Derivatives-Chain Rule

Chain Rule-Partial Derivatives!

Finding the partial derivatives using the chain rule:

Search Tags