How do you implicitly differentiate #7xy- 3 lny= 42#?

Question

Charles Anctil · Accepted Answer

#(dy/dx)=(-7y)/((7x) -(3/y))#

Ignore the #-3ln(y)# for now. Take the derivative of 7xy with product rule.

#(7x)(dy/dx)+(y)(7)#

#(7x)(dy/dx)+7y#

Now take a look at -3lny. Since 3 is a constant in front, you only need to worry about the derivaitve of lny and then multiply the 3 later.

#ln(y)=(1/y)(dy/dx)#

The derivative of a constant is always 0 so the equation will be equal to 0. Now combine your two parts together.

#(7x)(dy/dx)+7y -(3/y)(dy/dx)=0#

Subtract 7y on both sides.

#(7x)(dy/dx) -(3/y)(dy/dx)=-7y#

Then factor out #dy/dx# on the left side.

#(dy/dx)((7x) -(3/y))=-7y#

Isolate #dy/dx#.

#(dy/dx)=(-7y)/((7x) -(3/y))#

Emma Aldridge · Answer

undefined To implicitly differentiate the equation $7xy - 3 \ln(y) = 42$, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Apply the chain rule when differentiating terms involving $y$.
3. Solve for $\frac{dy}{dx}$ after differentiating each term.

Here's the step-by-step process:

$$\frac{d}{dx}(7xy) - \frac{d}{dx}(3 \ln(y)) = \frac{d}{dx}(42)$$
$$7\frac{d}{dx}(xy) - 3\frac{d}{dx}(\ln(y)) = 0$$
$$7\left(\frac{d}{dx}(x)y + x\frac{d}{dx}(y)\right) - 3\left(\frac{1}{y}\frac{dy}{dx}\right) = 0$$
$$7\left(y + x\frac{dy}{dx}\right) - 3\left(\frac{1}{y}\frac{dy}{dx}\right) = 0$$
$$7y + 7x\frac{dy}{dx} - \frac{3}{y}\frac{dy}{dx} = 0$$
$$7x\frac{dy}{dx} - \frac{3}{y}\frac{dy}{dx} = -7y$$
$$\frac{dy}{dx}(7x - \frac{3}{y}) = -7y$$
$$\frac{dy}{dx} = \frac{-7y}{7x - \frac{3}{y}}$$