ammu123
find the derivative of y with respect to x, t or theta.
(a) e^7 - 10x
Answer: -10e^7-10x
(b) 8xe^x - 8e^x
answer: 8xe^x
(c) y= (x^2 -2x+4)e^x
answer: (x^2+2)e^x
(d) y= sin e^theta^4
answer: (-4theta^3 e^-theta^4) cos e^-theta^4
please show the steps. thank you
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jim_thompson5910
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The first one is \[\Large e^{7-10x}\] right?
ammu123
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yes
jim_thompson5910
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The derivative of that isn't -10e^7-10x, so there has to be a typo somewhere
ammu123
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im not sure i think we have to use the u substitution method to find the answer.
jim_thompson5910
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oh wait, is the answer is \[\Large -10e^{7-10x}\] and not \[\Large -10e^{7}-10x\]
ammu123
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yea its the first one
jim_thompson5910
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this is why parenthesis are useful
if 7-10x is in the exponent, then say e^(7-10x)
jim_thompson5910
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ok that makes more sense now
ammu123
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sorry about that
jim_thompson5910
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let u = 7 - 10x, which means du/dx = -10
this means
\[\Large e^{7-10x}\]
turns into
\[\Large e^{u}\]
then derive to get
\[\Large e^{u}*\frac{du}{dx}\]
\[\Large e^{u}(-10)\]
\[\Large -10e^{u}\]
\[\Large -10e^{7-10x}\]
jim_thompson5910
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I'm using the chain rule, which is
If h(x) = f(g(x)), then
h ' (x) = f ' (g(x)) * g ' (x)
ammu123
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ok i just need help with the other 3 as well
jim_thompson5910
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whats the derivative of 8xe^x
ammu123
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im not sure
jim_thompson5910
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use the product rule
jim_thompson5910
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tell me what you get
ammu123
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is it 8xe^x
jim_thompson5910
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no, but that's part of it though
jim_thompson5910
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Product Rule:
if h(x) = f(x)*g(x), then
h ' (x) = f ' (x) * g(x) + f(x) * g ' (x)