Question

Find f′(a)f′(a). ?

f f(x)=7x^2−10x−35

Answer 1

`14a-10`

Explanation:

Method 1
From "first principles", i.e. from the definition of the derivative :

`f'(a) equiv lim_{h-> 0} (f(a+h)-f(a))/h`

Now, given `f(x) = 7x^2-10x-35`, we have

`f(a+h) = 7(a+h)^2-10(a+h)-35`
and
`f(a) = 7a^2-10a-35`

Thus

`f(a+h)-f(a) = 7{(a+h)^2-a^2}-10{a+h-a}-35+35`
`qquad = 7(2ah+h^2)-10h`

and so

`(f(a+h)-f(a))/h = {7(2ah+h^2)-10h}/h=14a+7h-10`

So, finally,

`f'(a) equiv lim_{h-> 0} (f(a+h)-f(a))/h = lim_{h to 0}(14a+7h-10)=14a-10`

Method 2

One can work the answer out with less effort if we use sonme standard properties of the derivative

`d/dx (f(x) pm g(x)) = d/dx f(x) pm d/dx g(x)`
`d/dx (cf(x)) = c d/dx f(x)`
`d/dx x^n = nx^(n-1)`

Using them,

#d/dx f(x) = d/dx (7x^2-10x-35) = d/dx(7x^2)-d/dx(10x)-d/dx(35)

`qquad = 7d/dxx^2-10 d/dx x^1 -35 d/dx x^0 = 7times 2x-10times 1-35times 0`
`qquad = 14x-10`

Thus

`f'(a) equiv d/dx f(x)|_{x=a} = 14a-10`