find derivative without using product//chain/or quotient rule..

Hello,

Please could someone run through deriving these 2 equations through algebraic expansion.

d/dx {sqrt(2y)}


and

d/dx {-1/((2x)^3)}

Thankyou in advance.

edit ... just figured out the second one..
the first one I still can't get my head around!

i'll run through what i'm doing ..

so,

d/dx[2y]^1/2

= 1/2*[2y]^-1/2

= 1/2*(1/sqrt(2y)

=1/(2*sqrt(2y)) ....

i should be getting 1/sqrt(2y) ... where am i going wrong?

Sqrt(2x)= sqrt(2)*sqrt(x)

Hello, euphmorning!

Quote:

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the first one . . .

So: .$\displaystyle \frac{d}{dx}\left[(2x)^{\frac{1}{2}}\right] $

. . . . $\displaystyle =\;\tfrac{1}{2}\cdot(2x)^{-\frac{1}{2}}{\color{red}\cdot 2} $

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Heya, thanks for the responses.. i'm afraid I still can't see how to arrive @ 1/sqrt(2y)

:/

The sqrt(2) is a constant so take the derivative of sqrt(x). 1/2(sqrt(x)). Multiply by the constant and arrive at sqrt(2)/2(sqrt(x)). Multiply that by sqrt(2)/sqrt(2) to get your answer.

Got it at last, thankyou very much.