differentiation with roots. (1 Viewer)

[Zak Ambrose]

having a bit od trouble with these 2 Qs.

f(x) = 1/4x^3

and

y= 6√ (x) - 1/(x^1/3)

 

[tacogym27101990]

having a bit od trouble with these 2 Qs.

f(x) = 1/4x^3

and

y= 6√ (x) - 1/(x^1/3)

is the first one f(x)= 1/(4x^3)
if it is write it as this
f(x)=(x^-3)/4 so
f'(x)= (-3x^-4)/4
= -3/(4x^4)

2.
y= 6(x^.5)-(x^-1/3) using the same index law as before
so y'= 6/2*(x^-.5) - -1/3*x^-4/3
=3/sqrtx +1/(x^4/3)

 

[Zak Ambrose]

yeh f(x)= 1/(4x^3)

i keep getting -12/(4x^4)

 

[Zak Ambrose]

how do i express 1/(x^4/3) in root form?

 

[Timothy.Siu]

how do i express 1/(x^4/3) in root form?

1/cuberoot(x^4)

 

[dim987654321]

always express these as indices so u can apply the chain rule when your differentiating.\
The two cases with roots so to speak is:

1. The first example with 1 as the numerator and some freaky root expression on the denominator. E.G. square root of (x - 5) I would do these by expressing the deno as an indice then as a negative indice (to get rid of the one in the numerator), then apply the chain rule.
2. You have roots on both numerator and denominator. Express both the denominator and the numerator as indices then apply the quotient rule, which will involve using the chain rule.

 

[Gigglydick]

You see Zak its relatively easy if you express as indices but if you dont. Crucial tonic will burn in hell

 

[Zak Ambrose]

You see Zak its relatively easy if you express as indices but if you dont. Crucial tonic will burn in hell

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and home address is 211 weir road lawrence, nsw.