Calculus help, please! Implicit differentiation with logarithms?

Question

I'm not sure how to approach this problem:

Use implicit differentiation to find dy/dx.
ln y − 7x ln x = −5

Any help would be appreciated!

Answer 1

ln(y) - 7xln(x) = -5



(1/y)(dy/dx) - 7x(1/x) - 7ln(x) = 0



(1/y)(dy/dx) - 7 - 7ln(x) = 0



(1/y)(dy/dx) = 7 + 7ln(x)



(1/y)(dy/dx) = 7(1 + ln(x))



(dy/dx) = 7y(1 + ln(x))

Answer 2

ln(y) - 7x ln(x) = -5



Implicit differentiation pretty much works by taking advantage of the chain rule. I mean,

the derivative of sin(x^5) is cos(x^5) * (5x^4).

The derivative of sin(x^3) = cos(x^3) * (3x^2), so by the same logic,

The derivative of sin(y) is cos(y) (dy/dx).

Got it? Every time you differentiate y, you write dy/dx.

Let's apply it to our equation.



ln(y) - 7x ln(x) = -5



So we will use the chain rule on ln(y), and the product rule on 7x ln(x). The derivative of -5 is just 0.



(1/y)(dy/dx) - 7(ln(x) + x(1/x)) = 0



Your next step is to solve for dy/dx.

Expanding it out a bit,



(1/y)(dy/dx) - 7ln(x) - 7x(1/x) = 0

(1/y)(dy/dx) - 7ln(x) - 7 = 0



Move everything without a dy/dx to the right hand side.



(1/y)(dy/dx) = 7ln(x) + 7

dy/dx = y ( 7ln(x) + 7 )

Answer 3

letting y ' = dy / dx:



y ' / y - (7 ln x + 7x / x) = 0

y ' / y - 7 ln x - 7 = 0

y ' / y = 7 ln x + 7

y ' = dy/dx = 7y (ln x + 1)



dy/dx = 7y (ln x + 1)

Answer 4

ln(y) - 7x * ln(x) = -5

dy / y - 7x * dx/x - 7 * ln(x) * dx = 0

dy/y - 7 * dx * (1 + ln(x)) = 0

dy/y = 7 * dx * (1 + ln(x))

dy/dx = 7 * y * (1 + ln(x))



Solve for y in the original equation



ln(y) = 7x * ln(x) - 5

y = e^(7x * ln(x) - 5)



dy/dx = 7 * e^(7x * ln(x) - 5) * (1 + ln(x))