How to use implicit differentiation on a trignometric and natural logarithm equation to find dy/dx

Question

For sin xy + cos y + x² = e^(x²) + ln x, how can I use implicit differentiation to find dy/dx ?

Answer 1

Implicit differentiation is treating the variable y as a function of x --> bascially using the definition of a function from algebra y = f(x). This means that f'(x) corresponds to the derivative of y with respect to x --> dy/dx. If you replace all the y's with f(x) you will see that there will be a bunch of places where the product rule or chain rule apply. This is the basis of implicit differentiation. After differentiating, we will re-replace f(x) and f'(x) back with y and dy/dx respectively.

For reference: Chain rule = deriv of outer function(inner function stays the same) * deriv of inner function

Product rule: d/dx g(x)*h(x) = g'(x) * h(x) + h'(x)*g(x)

sin xy + cos y + x² = e^(x²) + ln x

Replace y's with f(x)

sin(x * f(x)) + cos( f(x) ) + x^2 = e^(x^2) + ln(x)

For cleanliness I am going to differentiate the f(x) terms separately.

First term: Chain rule (outside function and inside function) and product rule (2 unique inside functions) both apply:

d/dx sin(x * f(x) ) = cos(x * f(x)) * (1* f(x) + x* f'(x) ) --> cos(x * f(x)) * ( f(x) + x*f'(x) )

REPLACE f(x) with y, and f'(x) with dy/dx:

cos(xy)* (y + x*dy/dx)

Second term: Chain rule (outside function and inside function) applies:

d/dx cos( f(x) ) = -sin(f(x)) * f'(x)

REPLACE f(x) with y, and f'(x) with dy/dx:

-sin(y) * dy/dx

Third, fourth, and fifth terms are only functions of x. e^(x^2) requires the chain rule while the remaining terms are basic derivatives:

d/dx x^2 = 2x

d/dx e^(x^2) = e^(x^2) * 2x --> 2x*e^(x^2)

d/dx lnx = 1/x

Put the equation back together:

cos(xy)* (y + x*dy/dx) + -sin(y) * dy/dx + 2x = 2xe^(x^2) + 1/x

Distribute the cos(xy) in the first term. The goal is to have all terms containing dy/dx on one side and all other terms on the other side:

cos(xy) * y + cos(xy) * x*dy/dx - sin(y)* dy/dx + 2x = 2xe^(x^2) + 1/x

x*cos(xy) * dy/dx - sin(y) *dy/dx = 2xe^(x^2) + 1/x - 2x - y*cos(xy)

Notice dy/dx is now the GCF of the terms on the left side. Pull out the GCF and rewrite what is left:

dy/dx * (xcos(xy) - sin(y)) = 2xe^(x^2) + 1/x - 2x - y*cos(xy)

The dy/dx can now be isolated because the leftover binomial can divide to the other side. Your final answer can sometimes look messy, and that is totally acceptable.

ANSWER:

dy / dx = [ 2xe^(x^2) + 1/x - 2x - y*cos(xy) ] / (xcos(xy) - sin(y))

Answer 2

sin(xy) + cosy + x² = e^{x^2} + lnx

cos(xy)[y + x(dy/dx)] - siny(dy/dx) + 2x = 2xe^{x^2} + 1/x

ycos(xy) + xcos(xy)(dy/dx) - siny(dy/dx) = 2xe^{x^2} + 1/x - 2x

(dy/dx)[xcos(xy) - siny] = 2xe^{x^2} + 1/x - ycos(xy) - 2x

dy/dx = [2xe^{x^2} + 1/x - ycos(xy) - 2x] / (xcos(xy) - siny)