How do you use implicit differentiation to find #y'# and evaluate #y'# at #(1,1)# for #xln(y) + 6y = 2x^3#?

1 Answer
Nov 11, 2017

Given: #xln(y) + 6y = 2x^3#

Differentiate each term with respect to x:

#(d(xln(y)))/dx + (d(6y))/dx = (d(2x^3))/dx#

Use the product rule, #(d(uv))/dx = (du)/dxv + u(dv)/dx#, on the first term, where #u = x# and and #v = ln(y)#:

#(d(xln(y)))/dx = (d(x))/dxln(y) + x(d(ln(y)))/dx#

On the second term, within the product rule, we must use the chain rule:

#(d(xln(y)))/dx = (d(x))/dxln(y) + x(d(ln(y)))/(dy)dy/dx#

#(d(xln(y)))/dx = ln(y) + x/ydy/dx#

Substitute this in place of the first term:

#ln(y) + x/ydy/dx + (d(6y))/dx = (d(2x^3))/dx#

For the next term, we use the chain rule:

#ln(y) + x/ydy/dx + (d(6y))/dy dy/dx = (d(2x^3))/dx#

#ln(y) + x/ydy/dx + 6dy/dx = (d(2x^3))/dx#

For the last term, we use the power rule #(d(x^n))/dx = nx^(n-1)#:

#ln(y) + x/ydy/dx + 6dy/dx = 6x^2#

Subtract #ln(y)# from both sides:

#x/ydy/dx + 6dy/dx = 6x^2- ln(y)#

Factor out #dy/dx#:

#(x/y + 6)dy/dx = 6x^2- ln(y)#

Divide by the leading coefficient:

#dy/dx = (6x^2- ln(y))/(x/y + 6)#

Multiply by #y/y#:

#dy/dx = (6yx^2- yln(y))/(x + 6y)#

Because the point #(1,1)# is not on the curve, I do not recommend that you evaluate the derivative at this point.