zeesbrat3
  • zeesbrat3
Determine the slope of the graph of x4 = ln(xy) at the point (1, e).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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freckles
  • freckles
find the slope by first differentiating both sides of the given equation
zeesbrat3
  • zeesbrat3
But hot can you find the derivative of e?
zeesbrat3
  • zeesbrat3
@freckles

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freckles
  • 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?
freckles
  • 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
zeesbrat3
  • 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.
zeesbrat3
  • zeesbrat3
So the first would be \[4x^3\] And the second would be \[\frac{ d }{ dx } \ln x + \ln y\]
freckles
  • freckles
oh okay so you used product rule for log but you still need to differentiate that sum: \[\frac{d}{dx}(\ln(x)+\ln(y)) \\ =\frac{d}{dx}\ln(x)+\frac{d}{dx}\ln(y)\] =?
zeesbrat3
  • zeesbrat3
\[\frac{ y }{ x }\]?
freckles
  • freckles
do you know that: where u is a function of x that: \[\frac{d}{dx}\ln(u)= \frac{u'}{u}\]
freckles
  • freckles
\[\frac{d}{dx}\ln(x)=\frac{(x)'}{x}=\frac{1}{x} \\ \frac{d}{dx}\ln(y)=\frac{y'}{y}\]
freckles
  • 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))
zeesbrat3
  • zeesbrat3
So \[4x^3 = 1 + \frac{ y' }{ e }\]
freckles
  • freckles
close you need to replace all the x's with 1
zeesbrat3
  • zeesbrat3
Ooo whoops! \[4 = 1 + \frac{ y' }{ }\]
zeesbrat3
  • zeesbrat3
theres supposed to be an e under the y prime
freckles
  • 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)
zeesbrat3
  • zeesbrat3
So \[3e = y' ?\]
freckles
  • freckles
yep
zeesbrat3
  • zeesbrat3
Thank you! Would you mind helping me with one more? Logs through me for a loop
freckles
  • freckles
sure
zeesbrat3
  • zeesbrat3
Thanks! \[\frac{ d }{ dx }[x^3e^x] = x^2e^x(3x+2)\]
zeesbrat3
  • zeesbrat3
I have to find whether that is true or not
freckles
  • freckles
well do you know product rule?
freckles
  • freckles
x^3e^x is a product of x^3 and e^x
freckles
  • 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: \[\frac{d}{dx}x^3=? \\ \frac{d}{dx}e^x=?\]
zeesbrat3
  • zeesbrat3
Right. So the first is \[3x^2\] and the second is 0?
freckles
  • freckles
the second is not 0 e^x is not a constant
freckles
  • freckles
for example e^2 and e^3 aren't the same so there is no way e^x remains the same
freckles
  • freckles
the derivative of e^x is e^x
zeesbrat3
  • zeesbrat3
Because it is a constant?
freckles
  • 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)

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