## zeesbrat3 one year ago Determine the slope of the graph of x4 = ln(xy) at the point (1, e).

1. freckles

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

2. zeesbrat3

But hot can you find the derivative of e?

3. zeesbrat3

@freckles

4. freckles

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?

5. freckles

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

6. zeesbrat3

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.

7. zeesbrat3

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

8. freckles

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

9. zeesbrat3

$\frac{ y }{ x }$?

10. freckles

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

11. freckles

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

12. freckles

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

13. zeesbrat3

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

14. freckles

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

15. zeesbrat3

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

16. zeesbrat3

theres supposed to be an e under the y prime

17. freckles

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

18. zeesbrat3

So $3e = y' ?$

19. freckles

yep

20. zeesbrat3

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

21. freckles

sure

22. zeesbrat3

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

23. zeesbrat3

I have to find whether that is true or not

24. freckles

well do you know product rule?

25. freckles

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

26. freckles

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

27. zeesbrat3

Right. So the first is $3x^2$ and the second is 0?

28. freckles

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

29. freckles

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

30. freckles

the derivative of e^x is e^x

31. zeesbrat3

Because it is a constant?

32. freckles

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)