Question

Find dy/dx if y2 = x2 + sin xy​

Answer 1

Answer:

By knowing chain rule, one can easily solve this problem.

Answer 2

$\frac{dy}{dx}$ = $\frac{2x + ycos(xy)}{2y-xcos(xy)}$ is the differentiation of y² = x² + sinxy when calculus differentiates the process of locating a function's derivatives.

Given that,

We have to find the $\frac{dy}{dx}$ of y² = x² + sinxy

We know that,

What is differentiation?

Differentiation is the term used in calculus to describe the process of locating a function's derivatives. A derivative is the pace at which one function changes in relation to another. Sir Isaac Newton set the foundations for the laws of differential calculus. Numerous scientific disciplines make use of the limits and derivatives ideas. The essential ideas of calculus are differentiation and integration.

Take the expression

y² = x² + sinxy

Differentiating on both sides

$\frac{d}{dx}$(y²) = $\frac{d}{dx}$(x²) + $\frac{d}{dx}$(sinxy)

2y $\frac{dy}{dx}$ = 2x + cos(xy) $\frac{d}{dx}$(xy)

2y $\frac{dy}{dx}$ = 2x + cos(xy) (y+x$\frac{dy}{dx}$)

2y $\frac{dy}{dx}$ = 2x + y cos(xy) +xcos(xy)$\frac{dy}{dx}$

2y $\frac{dy}{dx}$ -xcos(xy)$\frac{dy}{dx}$ = 2x + y cos(xy)

$\frac{dy}{dx}$(2y-xcos(xy) = 2x + y cos(xy)

$\frac{dy}{dx}$ = $\frac{2x + ycos(xy)}{2y-xcos(xy)}$

Therefore, $\frac{dy}{dx}$ = $\frac{2x + ycos(xy)}{2y-xcos(xy)}$ is the differentiation of y² = x² + sinxy when calculus differentiates the process of locating a function's derivatives.

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