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Find the partial derivative of the dependent variable or function with respect to each of the independent variables.$$f(x, y)=\frac{\tan ^{-1} 4 y}{2+x^{2}}$$

$f_{y}(x, y)=\frac{4}{\left(2+x^{2}\right)\left(1+16 y^{2}\right)}$

Calculus 3

Chapter 29

Partial Derivatives and Double Integrals

Section 3

Partial Derivatives

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Yeah, we want to find the first partial derivatives of the function F of X. Y is equal to the arc tangent of four Y divided by two plus X square. This question is testing our understanding of differentiation both from the techniques we learned in single variable calculus and for the purpose of partial derivatives and Multivariate calculus. The first part of it is we're searching for our DF DX and DY to obtain each. We use standard differentiation taking from single variable calculus. Where for DF dX we differentiate with respect to X treating any white terms the constant and similarly for the FBI we differentiate with respect to Y treating any X of constant. So the only white terms are temper why? The only X terms are two plus X squared. So FX is arc tanne for Y times negative. 1/2 plus X squared times two X. This is using the chain rule on the denominator. Before we proceed to FY, Which I have written at the bottom here, we should note that because we use the chain rule denominator, this two plus X square needs to be square rooted or rather squared in the denominator via the chain rule or the power rule chain rule. Then, for Fy below, we note that the only term is our 10. For why? So are derivative is 1/1 over two plus X squared. This was a constant. Remember times 4/1 plus 16. 1 square. The directive are 10 for why? Or for over two plus X squared times one for 16. Why? It's weird.

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