Question

If, 

f(x) = 7x+7–4x^2

then find the derivative f'(1).

Answer 1

The function provided in the question is given by,

f(x) =7x+7–4x^2

We know that, the formula for derivative is given as

d(x^n)/dx = nx^n–1

Now, differentiating f(x) with respect to x we get, 

df(x)/dx = d/dx[7x+7–4x^2]

=d(7x)/dx + d(7)/dx –d(4x^2)/dx

f'(x) = 7 + 0–8x

(since, the derivative of constant is 0.)

f'(x) =7–8x

On substituting x=1 in above derivative we get, 

f'(1)= 7–8(1)

f'(1) = –1

 

 

Hence the derivative f'(1) of a function f(x) =7x+7–4x^2 is equal to –1.