How do you find the derivative of #e^y+e^xsin(y)=sin(xy)#?

Answer 1

#dy/dx=(ycos(xy)-e^xsiny)/(e^y +e^xcosy-xcos(xy))#

#e^y+e^x siny=sin(xy)#
#e^y dy/dx+e^xsiny+e^xcosy dy/dx=ycos(xy)+xcos(xy)dy/dx#
#e^y dy/dx+e^xcosy dy/dx-xcos(xy)dy/dx=ycos(xy)-e^xsiny#
#dy/dx(e^y +e^xcosy-xcos(xy))=ycos(xy)-e^xsiny#
#dy/dx=(ycos(xy)-e^xsiny)/(e^y +e^xcosy-xcos(xy))#
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Answer 2

To find the derivative of the given equation (e^y + e^x \sin(y) = \sin(xy)) with respect to (x), we use implicit differentiation. Implicit differentiation involves taking the derivative of both sides of the equation with respect to (x), and then solving for (\frac{dy}{dx}) (the derivative of (y) with respect to (x)).

Given: [e^y + e^x \sin(y) = \sin(xy)]

Differentiate both sides with respect to (x), using the chain rule and the product rule where necessary.

  1. Differentiating (e^y) with respect to (x): Since (y) is a function of (x), by the chain rule, the derivative is (e^y \frac{dy}{dx}).

  2. Differentiating (e^x \sin(y)) with respect to (x): Using the product rule, we get (e^x \cos(y) \frac{dy}{dx} + e^x \sin(y)).

  3. Differentiating (\sin(xy)) with respect to (x): Using the chain rule and product rule, we get (\cos(xy) \cdot (y + x\frac{dy}{dx})).

Setting the derivatives equal gives: [e^y \frac{dy}{dx} + e^x \cos(y) \frac{dy}{dx} + e^x \sin(y) = \cos(xy) \cdot (y + x\frac{dy}{dx})]

Now, we solve for (\frac{dy}{dx}): [e^y \frac{dy}{dx} + e^x \cos(y) \frac{dy}{dx} = \cos(xy) \cdot (y + x\frac{dy}{dx}) - e^x \sin(y)]

Factor out (\frac{dy}{dx}) from the left side: [\frac{dy}{dx}(e^y + e^x \cos(y)) = \cos(xy) \cdot y + x\cos(xy)\frac{dy}{dx} - e^x \sin(y)]

Move terms involving (\frac{dy}{dx}) to one side: [\frac{dy}{dx}(e^y + e^x \cos(y)) - x\cos(xy)\frac{dy}{dx} = \cos(xy) \cdot y - e^x \sin(y)]

Factor out (\frac{dy}{dx}): [\frac{dy}{dx}(e^y + e^x \cos(y) - x\cos(xy)) = \cos(xy) \cdot y - e^x \sin(y)]

Finally, solve for (\frac{dy}{dx}): [\frac{dy}{dx} = \frac{\cos(xy) \cdot y - e^x \sin(y)}{e^y + e^x \cos(y) - x\cos(xy)}]

This is the derivative of (y) with respect to (x) for the given equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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