Question

Determine the derivative\(\frac{{dz}}{{dt}}\) with the help of the chain rule. The functions are\(z = {x^2} + {y^2} + xy,x = \sin t\) and\(y = {e^t}{\rm{. }}\)

Short answer

The derivative\(\frac{{dz}}{{dt}}\) with the help of chain rule is\((2x + y)\cos t + (2y + x){e^t}.\)

Step-by-step solution

Formula of chain rule.

The chain rule says:

\(\frac{d}{{dx}}(f(g(x))) = {f^\prime }(g(x)){g^\prime }(x)\)

It tells us how to differentiate composite numbers.

Use the chain rule for calculation.

The functions are\(z = {x^2} + {y^2} + xy,x = \sin t\) and\(y = {e^t}{\rm{. }}\)

With the help of chain rule, differentiate\(z\) with respect to\(t\) as follows:

\(\begin{aligned}{l}\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}}\\ = \frac{\partial }{{\partial x}}\left( {{x^2} + {y^2} + xy} \right)\frac{{d(\sin t)}}{{dt}} + \frac{\partial }{{\partial x}}\left( {{x^2} + {y^2} + xy} \right)\frac{d}{{dt}}\left( {{e^t}} \right)\\ = (2x + y)\cos t + (2y + x){e^t}\end{aligned}\)

Therefore, the derivative\(\frac{{dz}}{{dt}}\) with the help of chain rule is\((2x + y)\cos t + (2y + x){e^t}.\)