Question

Find the derivative of f(x) = 9x + 5 at x = 7.

Answer 1

The derivative of a variable raised to exponent of n, x^n is equal to the product of n and x raised to exponent of n-1.
                                   d(x^n) = (n)(x^n-1)
So, from the given
                                 d(9x + 5) = (1)(9x^1-0)  + 5(0)
                                                = 9
The given value of x is irrelevant to this item because the derivative does not contain any variable x. 

Answer 2

Answer:

The derivative of f(x) = 9x + 5 at x = 7 is:

                  $f'(7)=9$

Step-by-step explanation:

The derivative of a function is the rate of change of the dependent variable i.e. y=f(x) with respect to the independent variable i.e. x.

The derivative of the function f(x) is given by:

$f'(x)=\dfrac{df}{dx}$

Now, we know that:

$\dfrac{d}{dx}x=1\\\\\\and\\\\\\\dfrac{d}{dx}c=0$

where c is a constant term.

Here we have the function f(x) as:

$f(x)=9x+5$

Hence,

$\dfrac{d}{dx}f(x)=\dfrac{d}{dx}(9x)+\dfrac{d}{dx}5\\\\i.e.\\\\\dfrac{d}{dx}f(x)=9\dfrac{d}{dx}x+0\\\\\dfrac{d}{dx}f(x)=9$

i.e.  $f'(x)=9$

Also, when x =7 we have:

$f'(7)=9$