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