Calculus Homework Help!?

Question

Compute f '(a) algebraically for the given value of a

f(x)=-3/x; a=4 so f ' (x)=_______

f(x)=mx+b; a=19 so f ' (x)=_______



Compute f '(x) algebraically

f(x)=-6x+7 so f ' (x)=______

f(x)=8x-4x^2 so f ' (x)=_______

f(x)=3x^3+x so f ' (x)=______

Help me get the answers, and explain how you go it.

First one gets "Best Answer" points

Answer 1

f(x)=-3/x; a=4 so f ' (x)=3/x² ►►►►f ' (a)= 3/(4)² = 1/16
f(x)=mx+b; a=19 so f ' (x)=m►►►►f ' (a)= m = m



f(x)=-6x+7 so f ' (x)=6

f(x)=8x-4x² so f ' (x)= 8-8x

f(x)=3x³+x so f ' (x)= 9x²+1

Answer 2

f(x) = -3/x = -3 x^-1, so f'(x) = -3 (-1)x^(-2) = 3/x^2
f'(a) = 3/4^2 = 3/16.

f(x) = mx + b: f'(x) = m so f'(a) = m.

f(x) = -6x + 7: f'(x) = -6.

f(x) = 8x - 4x^2: f'(x) = 8 - 4(2x) = 8 - 8x

f(x) = 3x^3 + x: f'(x) = 3 (3x^2) + 1 = 9x^2 + 1

All these are simple applications of the power rule: d/dx x^n = n x^(n-1).

Answer 3

Hi
Just a short definition:
f'(x) is the tangent to f(x) at the point P(x|f(x))
f'(x) is called the derivative.

However, there is a nice simple rule applicable to most equations:
The derivative of any exponential function x^a+c is a*x^(a-1)
As you can see, the constant c falls away.

Here as an example the result of your last task:
f'(x)=9*x^2+1

Hope you got it :)

Answer 4

haha good luck with that! my brother is in the same boat as you with the calculus stuff.. looks difficult
sorry i was no help at all lol