What is the derivative of #e^y cos x = 3 + sin(xy)#?

Question

Charles Anctil · Accepted Answer

#dy/dx=(e^ysinx+ycos(xy))/(e^ycosx-xcos(xy))# #"differentiate " e^ycosx" using "color(blue)"product rule"# #"Given " f(x)=g(x)h(x)" then "# #f'(x)=g(x)h'(x)+h(x)g'(x)# #"differentiate " sin(xy)" using "color(blue)" chain rule"# #"Given " f(x)=g(h(x))" then"# #f'(x)=g'(h(x)).h'(x)# #"differentiate "color(blue)"implicitly with respect to x"# #rArre^y(-sinx)+cosxe^y.dy/dx=cos(xy).d/dx(xy)# #rArre^ycosx.dy/dx-e^ysinx=cos(xy)[x.dy/dx+y]# #rArre^ycosxdy/dx-xcos(xy)dy/dx=ycos(xy)+e^ysinx# #rArr dy/dx(e^ycosx-xcos(xy))=e^ysinx+ycos(xy)# #rArr dy/dx=(e^ysinx+ycos(xy))/(e^ycosx-xcos(xy))#

Ezra Adams · Answer

undefined To find the derivative of the equation $ e^y \cos(x) = 3 + \sin(xy) $ with respect to $ x $, we'll use implicit differentiation.

Differentiating both sides of the equation with respect to $ x $:

$$
\frac{d}{dx} \left( e^y \cos(x) \right) = \frac{d}{dx} \left( 3 + \sin(xy) \right)
$$

Using the chain rule and product rule on the left side and the product rule on the right side:

$$
\frac{d}{dx} \left( e^y \cos(x) \right) = e^y \frac{dy}{dx} \cos(x) - e^y \sin(x)
$$
$$
\frac{d}{dx} \left( 3 + \sin(xy) \right) = \cos(xy) \frac{dy}{dx} + y\cos(xy)
$$

Now, we'll rearrange the terms and solve for $ \frac{dy}{dx} $:

$$
e^y \frac{dy}{dx} \cos(x) - e^y \sin(x) = \cos(xy) \frac{dy}{dx} + y\cos(xy)
$$

$$
e^y \frac{dy}{dx} \cos(x) - \cos(xy) \frac{dy}{dx} = e^y \sin(x) + y\cos(xy)
$$

$$
\left( e^y \cos(x) - \cos(xy) \right) \frac{dy}{dx} = e^y \sin(x) + y\cos(xy)
$$

$$
\frac{dy}{dx} = \frac{e^y \sin(x) + y\cos(xy)}{e^y \cos(x) - \cos(xy)}
$$

So, the derivative of $ e^y \cos(x) = 3 + \sin(xy) $ with respect to $ x $ is $ \frac{e^y \sin(x) + y\cos(xy)}{e^y \cos(x) - \cos(xy)} $.