Just making sure im doing things right. suppose that
f(x)=2x^2+1/�ãx+1/4 x^3-x+1 Caclulate
f '(1)
So far i have 4x+(�ãx^-1)+3/4 x^2-1+0
then i substituted my 1 in and came out with 4 3/4
what is �ã ?
also, add some parentheses. It's not clear what f(x) is
sorry about that a shouldnt b in there it should be
f(x)=(2x^2)+(1/sqrt(x))+[(1/4)x^3]-x+1
the a thing was supposed to be a sqrt but i guess it doesnt copy over from work
Ok
f(x) = 2x^2 + 1/√x + (1/4)x^3 - x + 1
= 2x^2 + x^(-1/2) + (1/4)x^3 - x + 1
f'(x) = 4x - (1/2)x^(-3/2) + (3/4)x^2 - 1
f'(1) = 4 - (1/2)(1) + 3/4 - 1
= 13/4
I could be wrong but im not sure i came up with 1/4
To calculate f'(1), you need to find the derivative of f(x) first.
The given function is:
f(x) = 2x^2 + 1/(√x) + 1/4x^3 - x + 1
Taking the derivative of each term separately:
f'(x) = (d/dx)(2x^2) + (d/dx)(1/(√x)) + (d/dx)(1/4x^3) - (d/dx)(x) + (d/dx)(1)
Differentiating each term:
1. (d/dx)(2x^2) = 4x
2. (d/dx)(1/(√x)) = -1/(2√x^3) = -1/(2x√x)
3. (d/dx)(1/4x^3) = -3/(4x^4)
4. (d/dx)(x) = 1
5. (d/dx)(1) = 0
Now substitute x = 1 into f'(x):
f'(1) = 4(1) - 1/(2(1)(√1)) - 3/(4(1)^4) - 1 + 0
= 4 - 1/2 - 3/4 - 1
= 4 - 1/2 - 3/4 - 4/4
= 4 - 1/2 - 7/4
= 8/2 - 1/2 - 7/4
= (8 - 1 - 7)/4
= 0/4
= 0
Therefore, f'(1) = 0.