Question

If f(x) = 9x2 + 2x , then f ′(–1) is

Answer 1

Step-by-step explanation:

$let \: \: y = f(x) \\ \\ y = 9 {x}^{2} + 2x \\ \\ {y}^{. } = 18x + 2 \\ \\ put \: x = - 1 \\ \\ {y}^{.} = - 18 + 2 = - 16$

here's the solution for the given problem

Answer 2

If f(x) = 9x² + 2x then f ′(–1) is - 16

Given :

The function f(x) = 9x² + 2x

To find :

The value of f ′(–1)

Solution :

Step 1 of 3 :

Write down the given function

Here the given function is

f(x) = 9x² + 2x

Step 2 of 3 :

Find first order derivative

Differentiating both sides with respect to x we get

$\displaystyle \sf{ f'(x) = \frac{d}{dx} (9 {x}^{2} + 2x)}$

$\displaystyle \sf{ \implies f'(x) = \frac{d}{dx} (9 {x}^{2} ) +\frac{d}{dx} ( 2x)}$

$\displaystyle \sf{ \implies f'(x) = 9 \frac{d}{dx} ( {x}^{2} ) +2\frac{d}{dx} ( x)}$

$\displaystyle \sf{ \implies f'(x) = (9 \times 2x) +(2 \times 1)}$

$\displaystyle \sf{ \implies f'(x) = 18x + 2}$

Step 3 of 3 :

Find the value of f ′(–1)

Putting x = - 1 we get

$\displaystyle \sf{ f'( - 1) = 18 \times ( - 1)+ 2}$

$\displaystyle \sf{ \implies f'( - 1) = - 18 + 2}$

$\displaystyle \sf{ \implies f'( - 1) = - 16}$

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