Respuesta :

meerkat18
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:

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

                  [tex]f'(7)=9[/tex]

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:

[tex]f'(x)=\dfrac{df}{dx}[/tex]

Now, we know that:

[tex]\dfrac{d}{dx}x=1\\\\\\and\\\\\\\dfrac{d}{dx}c=0[/tex]

where c is a constant term.

Here we have the function f(x) as:

[tex]f(x)=9x+5[/tex]

Hence,

[tex]\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[/tex]

i.e.  [tex]f'(x)=9[/tex]

Also, when x =7 we have:

[tex]f'(7)=9[/tex]

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