How do you implicitly differentiate #7xy- 3 lny= 4/x#?

Question

Aurora Alvarado · Accepted Answer

#dy/dx=-(7x^2y^2+4y)/(7x^3y-3x^2)#

Find the derivative of each part.

Use product rule:

#d/dx(7xy)=7y+7xy'#

Recall that #d/dx(lnu)=(u')/u# through the chain rule:

#d/dx(-3lny)=-(3y')/y#

Rewrite #4/x# as #4x^-1#.

#d/dx(4x^-1)=-4x^-2=-4/x^2#

Add all these derivatives to find the complete differentiated expression, then solve for #y'#.

#7y+7xy'-(3y')/y=-4/x^2#

Multiply everything by #x^2y# to clear fractions.

#7x^2y^2+7x^3y(y')-3x^2y'=-4y#

#y'(7x^3y-3x^2)=-7x^2y^2-4y#

#y'=-(7x^2y^2+4y)/(7x^3y-3x^2)=dy/dx#

Emilia Aguilar · Answer

undefined To implicitly differentiate $7xy - 3\ln(y) = \frac{4}{x}$, follow these steps:

1. Differentiate each term with respect to $x$.
2. Apply the chain rule when differentiating $\ln(y)$.

Differentiating each term:

$$\frac{d}{dx}(7xy) - \frac{d}{dx}(3\ln(y)) = \frac{d}{dx}\left(\frac{4}{x}\right)$$

Apply product rule to $7xy$:

$$7\frac{d}{dx}(xy) + x\frac{d}{dx}(7y) - 3\frac{d}{dx}(\ln(y)) = \frac{d}{dx}\left(\frac{4}{x}\right)$$

Apply chain rule to $\ln(y)$:

$$7\left(\frac{dy}{dx}y + x\frac{dy}{dx}\right) - 3\left(\frac{1}{y}\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{4}{x}\right)$$

Differentiate $\frac{4}{x}$:

$$-\frac{4}{x^2} = -\frac{4}{x^2}$$

Simplify and solve for $\frac{dy}{dx}$:

$$7xy' + 7y + xy' - \frac{3}{y}y' = -\frac{4}{x^2}$$

$$7xy' + xy' - \frac{3}{y}y' = -\frac{4}{x^2} - 7y$$

$$xy'(7 + 1) - \frac{3}{y}y' = -\frac{4}{x^2} - 7y$$

$$8xy' - \frac{3}{y}y' = -\frac{4}{x^2} - 7y$$

$$y'(8x - \frac{3}{y}) = -\frac{4}{x^2} - 7y$$

$$y' = \frac{-\frac{4}{x^2} - 7y}{8x - \frac{3}{y}}$$

This is the implicit derivative of the given function.