# Thread: finding the differiential of the function

1. ## finding the differiential of the function

Can anyone teach me step by step how to differetiate?

1) f(x)=2x^3+x

2)f(x)=3/sqrt(x)

3)f(x)=3x^5/6+7x^2/3

2. do you remember the power rule?

$\frac d{dx}x^n = nx^{n - 1}$

if there is a constant multiplying, it doesn't matter, just take it along

$\frac d{dx}cx^n = c \frac d{dx}x^n = cnx^{n - 1}$ for c a constant
Originally Posted by lemontea
Can anyone teach me step by step how to differetiate?

1) f(x)=2x^3+x
apply the power rule for each term

2)f(x)=3/sqrt(x)
note that this is the same as $f(x) = 3x^{-1/2}$ and apply the power rule

3)f(x)=3x^5/6+7x^2/3
i am not sure what you wrote here. please use parentheses to clarify

type fractions like (numerator)/(denominator)

example, type (3x + 1)/(x - 2) to mean $\frac {3x + 1}{x - 2}$

3. I'm assuming you dont need the proof of this from first principles,we would be using the rules that

$\frac{d}{dx}(ax^n)=anx^{n-1}$

and that

$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d }{dx}(g(x))$

for the first one we would have in the first term the power (n) is 3 and the coefficient before it (a) is 2 so the differential of that would be $2 \times 3 x^2=6x^2$

then the second term a=1 and n=1 so the differential of that would be 1 giving us

$f'(x)=6x^2+1$

can you do the same for the other parts now in the second remember that $\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$