Implicit differentiation: ycosx=x^2+y^2

2 answers:

kramer3 years ago

6 0

Y cos x = x^2 + y^2

Taking derivative:
<span>
y' cos x - y sin x = 2x + 2y y' </span>

Simplifying:

<span>y' = (- 2x - y sin x)/(2y - cos x)</span>

mixas84 [53]3 years ago

5 0

Answer:

Step-by-step explanation:

Given is an implicit function of y in x.

$ycos x = x^2+y^2\\$

The steps shown were the differentiation on both the sides

Left side done using product rule as well as chain rule and right side addition rule and chain rule

The steps got as

$((y)(-sinx)+(cosx)(dy/dx)=2x+2y(dy/dx)$ is correct

Next step is to group all dy/dx terms together

$cosx(dy/dx) - 2y(dy/dx) = ysinx+2x$ is also right step

Now rewrite left side as

$\frac{dy}{dx} (cosx-2y)=2x+ysinx\\\frac{dy}{dx}=\frac{ysinx+2x}{cosx-2y}$

Thus both the steps and answer are correct.

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