Question

f(x)=5x^4-(1/3x^2)+ax f'(x)=?

Answer 1

Differentiation of f(x) with respect to x gives f'(x)
Is ur question
\[\frac{5x^4-1}{3x^2+ax}\]?

Answer 2

no no

Answer 3

1/3(x)

Answer 4

\(f(x)=bx^n\)
\(f'(x)=b*nx^{n-1}\)
does this help?

Answer 5

i would have to apply on all those terms

Answer 6

f(x)=b*x^n. this is similar to f(x)=5x^4
in place of b u have 5 and n is 4.
f'(x) of bx^n is b*nx^{n-1}. so can u find f'(x) of 5x^4.

Answer 7

u can use it for other terms (-1/3)x^2 and +ax. Can u find @jadi?

Answer 8

i would have to apply more then one time ???

Answer 9

yup. hav u found the differentiation of 5x^4?

Answer 10

20x^3

Answer 11

ya. very good:) similarly find differentiation of (-1/3)x^2 and +ax and combine them to find the total differentiation of f(x)=5x^4-(1/3x^2)+ax

Answer 12

@jadi hav u found?

Answer 13

it 0

Answer 14

noo. what is the differentiation of ax?

Answer 15

a

Answer 16

ya good:) and diffrentiation of (-1/3)x^2?

Answer 17

2/3x

Answer 18

Guess u forgot the - sign. Now add them all together to find the differentiation of 5x^4-(1/3x^2)+ax. Can u do @jadi?

Answer 19

no i could not

Answer 20

f(x)=5x^4-(1/3x^2)+ax
f'(x)=20x^3-(2/3)x+a
Getting it?

Answer 21

ja is that all

Answer 22

yup:)

Answer 23

ooo thank u very much

Answer 24

welcome:)