IsTim
  • IsTim
Express each function as a power with a rational exponent, and then differentiate. Express each answer using positive exponents. y=sqrt(2x-3x^5)
Mathematics
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SOLVED
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katieb
  • katieb
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IsTim
  • IsTim
Well. I'm just doing the first part. I got 1/2(2-15x^4)^-1/2
dumbcow
  • dumbcow
ok sqrt is same as exponent of 1/2 then apply chain rule to differentiate good..wait don't change inside...multiply whole thing by derivative of inside
dumbcow
  • dumbcow
\[(2-15x^{4})\frac{1}{2}(2x-5x^{3})^{-1/2}\]

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More answers

dumbcow
  • dumbcow
**typo 3x^5
IsTim
  • IsTim
In my textbook, it says: \[y=(2x-3x^5)^{1/2}\]
dumbcow
  • dumbcow
?? yes that is original function right
IsTim
  • IsTim
No, the answer. Well, the first half that isn't dy/dx.
dumbcow
  • dumbcow
oh yeah...sqrt is same as exponent of 1/2
IsTim
  • IsTim
But I don't know how to get \[dy/dx=(2=15x^4 )/(2\sqrt{2x-3x^5}\]
IsTim
  • IsTim
Also, why is their exponent positive, but ours is negative?
dumbcow
  • dumbcow
same answer i posted....ours is negative because its in numerator, when you flip it to denominator the sign of exponent changes
IsTim
  • IsTim
Ok. I understand the first part. Thanks. I'll call out if I need help on the second part.
IsTim
  • IsTim
Um, to find the derviative I use 1/2(2x−5x^3)−1/2 right.
dumbcow
  • dumbcow
whoah ...um we were doing 2nd part (finding derivative this whole time) or at least thats what i was doing
IsTim
  • IsTim
Oh no! How do I do the first part?
dumbcow
  • dumbcow
sqrt is same as exponent of 1/2
IsTim
  • IsTim
To get dy/dx=(2=15x^4)
IsTim
  • IsTim
/(22x−3x5−−−−−−−√
IsTim
  • IsTim
I guess I can't copy Latex....
IsTim
  • IsTim
dy/dx=(2=15x^4)/(2sqrt(2x−3x^5))
IsTim
  • IsTim
How do I figure that out?
dumbcow
  • dumbcow
let me try this, part 1 is super easy all you are doing is rewriting it using rational exponents \[\large \sqrt[n]{x} = x^{1/n}\] \[\sqrt{2x-3x^{5}} = (2x-3x^{5})^{1/2}\] part 2: find derivative using chain rule u = 2x-3x^5 du = 2-15x^4 \[\frac{d}{du} u^{1/2} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}\] multiply by du and substitute 2x-3x^5 back in for u \[\rightarrow \frac{2-15x^{4}}{2\sqrt{2x-3x^{5}}}\]
IsTim
  • IsTim
How'd you get du = 2-15x^4?
IsTim
  • IsTim
@dumbcow When you have time...
dumbcow
  • dumbcow
isn't that what you got in your very 1st post ^^ ?? anyway, its the derivative of 2x-3x^5 using power rule \[\large \frac{d}{dx} x^{n} = n*x^{n-1}\]
IsTim
  • IsTim
Ok. thanks.

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