Question

If $u=f(x,y)$ is a differentiable function of $x$ and $y$, where $x$ and $y$ are differentiable functions of $t$ then:

• A. $\frac{du}{dt}=\frac{\partial f}{\partial x}.\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}.\frac{\partial y}{\partial t}$
• B. $\frac{du}{dt}=\frac{\partial f}{\partial x}.\frac{dx}{dt}-\frac{\partial f}{\partial y}.\frac{\partial y}{\partial t}$
• C. $\frac{du}{dt}=\frac{\partial f}{\partial x}.\frac{dx}{dt}+\frac{\partial f}{\partial y}.\frac{dy}{dt}$
• D. $\frac{du}{dt}=\frac{\partial f}{\partial x}.\frac{\partial x}{\partial t}-\frac{\partial f}{\partial y}.\frac{\partial y}{\partial t}$

Answer 1

Answer: (C)

$\frac{du}{dt}=\frac{\partial f}{\partial x}.\frac{dx}{dt}+\frac{\partial f}{\partial y}.\frac{dy}{dt}$

Given $u=f(x,y)$ is a a differentiable function of $x$ and $y$, where $x,y$ are differentiable functions of $t$.

We have to find $\frac{du}{dt}$

Since $x,y$ are differentiable functions of $t$

Let $x=g\left(t\right),y=h\left(t\right)$

Thus by chain rule we get

$\frac{du}{dt}=\frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial f}{\partial y}\cdot \frac{dy}{dt}$