Question

Determine the slope of the graph of x4 = ln(xy) at the point (1, e).

Answer 1

find the slope by first differentiating both sides of the given equation

Answer 2

But hot can you find the derivative of e?

Answer 3

@freckles

Answer 4

why do you want the derivative of e?
by the way e is a constant and derivative of e is 0.
But why do you need this information?

Answer 5

\[\frac{d}{dx}x^4=? \\ \frac{d}{dx}\ln(xy)=?\]
the first can be found by using power rule
second can be found using chain rule and product rule

Answer 6

Sorry, I'm just confused as to how to solve for the answer.. This question is on a practice test so I need to learn how to do it for the actual test.

Answer 7

So the first would be \[4x^3\]
And the second would be \[\frac{ d }{ dx } \ln x + \ln y\]

Answer 8

oh okay so you used product rule for log
but you still need to differentiate that sum:
\[\frac{d}{dx}(\ln(x)+\ln(y)) \\ =\frac{d}{dx}\ln(x)+\frac{d}{dx}\ln(y)\]
=?

Answer 9

\[\frac{ y }{ x }\]?

Answer 10

do you know that:
where u is a function of x
that:
\[\frac{d}{dx}\ln(u)= \frac{u'}{u}\]

Answer 11

\[\frac{d}{dx}\ln(x)=\frac{(x)'}{x}=\frac{1}{x} \\ \frac{d}{dx}\ln(y)=\frac{y'}{y}\]

Answer 12

\[x^4=\ln(xy) \\ \text{ after differentiating both sides we have } \\ 4x^3=\frac{1}{x}+\frac{y'}{y} \]
now you can replace x with 1 and y with e
and solve for y' (the slope at (1,e))

Answer 13

So \[4x^3 = 1 + \frac{ y' }{ e }\]

Answer 14

close you need to replace all the x's with 1

Answer 15

Ooo whoops! \[4 = 1 + \frac{ y' }{ }\]

Answer 16

theres supposed to be an e under the y prime

Answer 17

\[4=1+\frac{y'}{e}\]
yep now just solve for y' (this will be the slope at (1,e) since we replaced x with 1 and y with e)

Answer 18

So \[3e = y' ?\]

Answer 19

yep

Answer 20

Thank you! Would you mind helping me with one more? Logs through me for a loop

Answer 21

sure

Answer 22

Thanks!
\[\frac{ d }{ dx }[x^3e^x] = x^2e^x(3x+2)\]

Answer 23

I have to find whether that is true or not

Answer 24

well do you know product rule?

Answer 25

x^3e^x is a product of x^3 and e^x

Answer 26

\[\frac{d}{dx}(x^3e^x)=e^x \frac{d}{dx}x^3+x^3 \frac{d}{dx}e^x\]
what are the following:
\[\frac{d}{dx}x^3=? \\ \frac{d}{dx}e^x=?\]

Answer 27

Right. So the first is \[3x^2\] and the second is 0?

Answer 28

the second is not 0
e^x is not a constant

Answer 29

for example e^2 and e^3 aren't the same
so there is no way e^x remains the same

Answer 30

the derivative of e^x is e^x

Answer 31

Because it is a constant?

Answer 32

I thought I just said it wasn't a constant and gave a reason why it wasn't a constant.

e is a constant

e^x is not a constant (it varies) (Example: e^2 is not the same as e^3)