Question

Determine the slope of the graph of x^4=ln(xy) at the point (1,e)

Answer 1

It is 3e. Differentiating x^4=ln(xy), we have to implicitly differentiate this.

$4x^{3}= \frac{1}{xy} \cdot \frac{d}{dx}(xy)$

The derivative of xy you have to use the chain rule.

$\frac{d}{dx}(xy)=x \cdot \frac{dy}{dx}+y$

So now it's

$4x^{3}=\frac{y+\frac{dy}{dx} \cdot x}{xy}$

Now we want the slope when x=1 and y=e

$4(1)^{3}=\frac{e+\frac{dy}{dx}}{e}$

$4=\frac{e+\frac{dy}{dx}}{e}$

$4=1+\frac{dy}{dx} \cdot \frac{1}{e}$

$\frac{dy}{dx}=3e$

Answer 2

Answer:

Answer is 3e.

Step-by-step explanation:

The slope is given by the derivative at the given point.

Differentiating:-

4x^3 = 1/xy *(x * dy/dx + y)

4x^3 = 1/y * dy/dx + 1/x

dy/dx = (4x^3 - 1/x) * y

At the point (1, e) x = 1 and y = e so substituting

slope at (1,e) = dy/dx =  (4(1)^3 - 1/1) * e

= 3e  (answer)