How do you use implicit differentiation to find dy/dx given #e^(xsiny)=y#?

2 Answers
Oct 7, 2016

#e^(xsiny) = y#

Rewrite as:

#e^square = y#

Where #square = xsiny#.

Let's find the derivative of #square#.

#square' = 1 xx siny + x xx cosy(dy/dx)#

#square' = siny + xcosy(dy/dx)#

The derivative of #y = e^f(x)# is always #f'(x) xx e^(f(x))#, so #dy/dx = (siny + xcosy(dy/dx))e^(xsiny)#

However, to determine the complete derivative of the function, we need to isolate #dy/dx# on one side of the equation.

#dy/dx = sinye^(xsiny) + xcosy(dy/dx)(e^(xsiny))#

#dy/dx - xcosy(dy/dx)(e^(xsiny)) = sinye^(xsiny)#

#dy/dx(1 - xcosye^(xsiny)) = sinye^(xsiny)#

#dy/dx = (sinye^(xsiny))/(1 - xcosye^(xsiny))#

Substitute the initial function as #y# for #e^(xsiny)#.

#dy/dx = (ysiny)/(1 - xycosy)#

Hopefully this helps!

Oct 7, 2016

Please follow the instructions below...

Explanation:

#e^(xsiny)=y#

#lny=xsiny#

Now use both implicit differentiation formulas:

  1. On the left... #(dy)/(dy)*(dy)/(dx)=(dy)/(dx)#

  2. On the right... #f(x)g'(y)(dy)/(dx)+g(y)f'(x)#

If you do this, what you'll get is...

#1/y*(dy)/(dx)=x*cosy*(dy)/(dx)+siny#

#1/y*(dy)/(dx)-x*cosy*(dy)/(dx)=siny#

#(dy)/(dx){1/y-x*cosy}=siny#

#(dy)/(dx){1/y-(xycosy)/y}=siny#

#(dy)/(dx){(1-xycosy)/y}=siny#

#(dy)/(dx)=siny*y/(1-xycosy)#

#(dy)/(dx)=(ysiny)/(1-xycosy)#