How do you differentiate #sinx+xy+y^5=pi#?

Question

Christopher Agresta · Accepted Answer

#dy/dx=-(y+cosx)/(x+5y^4)# #sinx +xy +y^5=pi# Differentiating both sides with respect to #x# we get #cosx+(y+xdy/dx)+5y^4dy/dx=0# #:.dy/dx=-(y+cosx)/(x+5y^4)#

Charles Arocha · Answer

Given: #sinx+xy+y^5=pi#

Differentiate each term:

#(d(sinx))/dx+(d(xy))/dx+(d(y^5))/dx=(d(pi))/dx#

The first term is well known #(d(sinx))/dx = cos(x)#:

#cos(x)+(d(xy))/dx+(d(y^5))/dx=(d(pi))/dx#

The second term requires the use of the product rule:

#(d(xy))/dx = (d(x))/dxy+x(d(y))/dx#

#(d(xy))/dx = y+xdy/dx#

#cos(x)+y+xdy/dx+(d(y^5))/dx=(d(pi))/dx#

The third term requires the use of the chain rule:

#(d(y^5))/dx = (d(y^5))/dydy/dx#

#(d(y^5))/dx = 5y^4dy/dx#

#cos(x)+y+xdy/dx+5y^4dy/dx=(d(pi))/dx#

For the last term, we invoke the fact that the derivative of a constant is 0:

#cos(x)+y+xdy/dx+5y^4dy/dx=0#

Move all of the terms NOT containing #dy/dx# to the right:

#xdy/dx+5y^4dy/dx=-(cos(x)+y)#

Factor out #dy/dx# on the left:

#(x+5y^4)dy/dx=-(cos(x)+y)#

Divide both sides by the leading coefficient:

#dy/dx=-(cos(x)+y)/(x+5y^4)#

Axel Alvarez · Answer

undefined To differentiate the equation $ \sin(x) + xy + y^5 = \pi $ with respect to $ x $, we use implicit differentiation. Taking the derivative of each term individually, we get:

$ \frac{d}{dx}(\sin(x)) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^5) = \frac{d}{dx}(\pi) $

Using the chain rule and product rule where necessary:

$ \cos(x) + x\frac{dy}{dx} + y + 5y^4\frac{dy}{dx} = 0 $

Rearranging terms:

$ (x + 5y^4)\frac{dy}{dx} = -\cos(x) - y $

Finally, solving for $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = \frac{-\cos(x) - y}{x + 5y^4} $