Question

Suppose w=xy+yzw=xy+yz, where x=e2t, y=2+sin(4t)x=e2t, y=2+sin⁡(4t), and z=2+cos(5t)z=2+cos⁡(5t).
a. use the chain rule to find dwdtdwdt as a function of x, y, z, and t. do not rewrite x, y, and z in terms of t, and do not rewrite e2te2t as x.

Answer 1

$w(x,y,z)=xy+yz$
$x(t)=e^{2t}$
$y(t)=2+\sin4t$
$z(t)=2+\cos5t$

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

$\dfrac{\mathrm dw}{\mathrm dt}=y\dfrac{\mathrm dx}{\mathrm dt}+(x+z)\dfrac{\mathrm dy}{\mathrm dt}+y\dfrac{\mathrm dz}{\mathrm dt}$

$\dfrac{\mathrm dw}{\mathrm dt}=2ye^{2t}+4(x+z)\cos4t-5y\sin5t$

Answer 2

The chain rule is one of the techniques to take the derivative of an expression.

The value of the differentiation is: $\mathbf{\frac{dw}{dt} = 2ye^{2t} +4(x + z)cos(4t) -5ysin(5t)}$

The given parameters are:

$\mathbf{w = xy + yz}$

$\mathbf{x = e^{2t}}$

$\mathbf{y = 2 + sin(4t)}$

$\mathbf{z = 2 + cos(5t)}$

Expand

$\mathbf{w = y(x + z)}$

Differentiate w with respect to t using chain rule.

So, we have:

$\mathbf{w' = \frac{dw}{dx} \times x' +\frac{dw}{dy} \times y' +\frac{dw}{dz} \times z'}$

$\mathbf{w' = y \times x'+(x + z)\times y' +y \times z'}$

Differentiate x, y and z with respect to t

$\mathbf{x = e^{2t}}$

$\mathbf{x' = 2e^{2t}}$

$\mathbf{y = 2 + sin(4t)}$

$\mathbf{y' = 4cos(4t)}$

$\mathbf{z = 2 + cos(5t)}$

$\mathbf{z' = -5sin(5t)}$

Substitute these values in the equation of w'

$\mathbf{w' = y \times x'+(x + z)\times y' +y \times z'}$

$\mathbf{w' = y \times 2e^{2t} +(x + z) \times 4cos(4t) -y\times 5sin(5t)}$

$\mathbf{w' = 2ye^{2t} +4(x + z)cos(4t) -5ysin(5t)}$

Hence,

$\mathbf{\frac{dw}{dt} = 2ye^{2t} +4(x + z)cos(4t) -5ysin(5t)}$

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