I am revising for a calculus exam I have on Monday. One of the past paper questions on partial derivatives is given below:
Let $T(x,y) = \tan(xy^2) + 3y - 2xy$
i) Find the first order partial derivatives of $T$ with respect to $x$ and $y$. I did this and got:
$$\frac{\partial T(x,y)}{\partial x} = \sec^2(xy^2)y^2 - 2y$$ $$\frac{\partial T(x,y)}{\partial y} = \sec^2(xy^2)2yx + 3 - 2x$$
ii) Using the above calculate $\frac{dy}{d x}$: Now I am familiar with the product and chain rules for partial differentiation but because this is not a composition of functions, drawing the usual variable dependence diagram and summing up the products of those variables using this diagram doesn't seem to make any sense.
My question is therefore: How do I calculate $\frac{dy}{d x}$ from two partial derivatives?
I am sorry that I do not know LaTeX. Also, if you can, please make didactic concessions. I am a mere undergraduate and have only been studying calculus for two and a half months.
**Update: ** This question was asked in the exam. I got my calculus exam results a week ago and passed with distinction. I know I got this question right. :) Thank you all!
By using chain rule in multivariable calculus .... $dy/dx$ = $-f_x/f_y$