Home » mcq » Engineering maths » Calculus » Let w = f(x, y), where x and y are functions of t. Then, according to the chain rule, \[\frac{{{\rm{dw}}}}{{{\rm{dt}}}}\] is equal A. \[\frac{{{\rm{dw}}}}{{{\rm{dx}}}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + \frac{{{\rm{dw}}}}{{{\rm{dy}}}}\frac{{{\rm{dt}}}}{{{\rm{dt}}}}\] B. \[\frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{\partial {\rm{x}}}}{{\partial {\rm{t}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{\partial {\rm{y}}}}{{\partial {\rm{t}}}}\] C. \[\frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\] D. \[\frac{{{\rm{dw}}}}{{{\rm{dx}}}}\frac{{\partial {\rm{x}}}}{{\partial {\rm{t}}}} + \frac{{{\rm{dw}}}}{{{\rm{dy}}}}\frac{{\partial {\rm{y}}}}{{\partial {\rm{t}}}}\]
”[ rac{{{ m{dw}}}}{{{ m{dx}}}} rac{{{ m{dx}}}}{{{ m{dt}}}}
Answer is Right!
Answer is Wrong!
” option2=”\[\frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{\partial {\rm{x}}}}{{\partial {\rm{t}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{\partial {\rm{y}}}}{{\partial {\rm{t}}}}\]” option3=”\[\frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\]” option4=”\[\frac{{{\rm{dw}}}}{{{\rm{dx}}}}\frac{{\partial {\rm{x}}}}{{\partial {\rm{t}}}} + \frac{{{\rm{dw}}}}{{{\rm{dy}}}}\frac{{\partial {\rm{y}}}}{{\partial {\rm{t}}}}\]” correct=”option3″]
The correct answer is $\boxed{\frac{{\partial {\rm{w}}}}{{\partial {\rm{x}}}}\frac{{\partial {\rm{x}}}}{{\partial {\rm{t}}}} + \frac{{\partial {\rm{w}}}}{{\partial {\rm{y}}}}\frac{{\partial {\rm{y}}}}{{\partial {\rm{t}}}}}$.
The chain rule states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function evaluated at the point of composition. In this case, the outer function is $f$ and the inner function is $(x, y)$. Therefore, the derivative of $w$ with respect to $t$ is:
$$\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt}$$
where $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$ are the partial derivatives of $w$ with respect to $x$ and $y$, respectively, and $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are the derivatives of $x$ and $y$ with respect to $t$.
The other options are incorrect because they do not take into account the fact that $x$ and $y$ are functions of $t$.