#I actually have no idea on how to get the derivatives of this.

6 messages · Page 1 of 1 (latest)

rain wigeon
fickle sandBOT
#

I actually have no idea on how to get the derivatives of this.

buoyant bear
#

let's say you have y = sqrt(x),
Square root of something is that thing raised to 1/2
So y = x^(1/2)
Now to get derivative, use the rule power rule ( (x^n)' = nx^(n-1) )
So y'= (1/2) * x(1/2 - 1)
= (1/2)*x^(-1/2)
= (1/2) * 1/x^(1/2)

So for 3)
We can simplify it to be y = 2*sqrt(1/x) - sqrt(3)* sqrt(x)
= 2*(x^(-1/2)) - sqrt(3)*(x^(1/2))

#

For 4, you could take x^3 common and simplify

jovial bough
#

rationalize the denomnateros