Respuesta :
In this question, we find the derivatives, using the chain's rule.
Doing this, the derivative is:
[tex]\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}[/tex]
Chain Rule:
Suppose we have a function [tex]w(x,y,z), x = x(t), y = y(t), z = z(t)[/tex], and want to find it's derivative as function of t. It will be given by:
[tex]\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}[/tex]
Thus, we have to find the desired derivatives, which are:
- w of x:
[tex]\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}[/tex]
Considering [tex]x = \frac{1}{t}, y = t, z = t^3[/tex]
[tex]\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}[/tex]
- w of y:
[tex]\frac{dw}{dy} = -5x\cos{(xy)}[/tex]
Considering [tex]x = \frac{1}{t}, y = t[/tex]
[tex]\frac{dw}{dy} = -\frac{5}{t}\cos{1}[/tex]
- w of z:
[tex]\frac{dw}{dz} = -x\cos{(xz)}[/tex]
Considering [tex]x = \frac{1}{t}, z = t^3[/tex]
[tex]\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}[/tex]
- Derivatives of x, y and z as functions of t:
[tex]\frac{dx}{dt} = -\frac{1}{t^2}[/tex]
[tex]\frac{dy}{dt} = 1[/tex]
[tex]\frac{dz}{dt} = 3t^2[/tex]
- Derivative of w as function of t.
Now, we just replace what we found into the formula. So
[tex]\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}[/tex]
[tex]\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2[/tex]
Applying the multiplications:
[tex]\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}[/tex]
Applying the simplifications:
[tex]\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}[/tex]
Which is the derivative.
For more on the chain rule, you can check https://brainly.com/question/12795383
