Question

Evaluate each of the following functions, find the indicated derivative

1. 𝑓(𝑥) = 𝑥^2 − 4𝑥 + 1; 𝑓′(2)

2. 𝑓(𝑥) = 𝑥^3 + 2; 𝑓′(−2)

3. 𝑓(𝑥) = 2𝑥^4 + 3𝑥^3 − 2𝑥 + 7; 𝑓′(0)

4. f(x) = √2x + 7; f'(1)

5. f(x) = 1 + √x^2 + 3x +6; f'(2)

6. f(x) = x/x+4 ; f'(-3)

7. f(x) = x^2 +3/4 - x^2 ; f'(-1)

8. f(x) = √x + 3/x - 4 ;f'(1)

9. f(x) = 3- √ 5x - 9/2x - 1 f'(2)

10. f(x) = 2x^2 + 3 - 1; f'(-1)

Answer 1

Solution (1)

$f\left( x \right) = {x^2} - 4x + 1\\ f'\left( x \right) = 2x - 4\\ f'\left( 2 \right) = 2\left( 2 \right) - 4 = 0$

Solution (2)

$f\left( x \right) = {x^3} + 2\\ f'\left( x \right) = 3{x^2}\\ f'\left( { - 2} \right) = 3{\left( { - 2} \right)^2} = 12$

Solution (3)

$f\left( x \right) = 2{x^4} + 3{x^3} - 2x + 7\\ f'\left( x \right) = 8{x^3} + 9{x^2} - 2\\ f'\left( 0 \right) = 8{\left( 0 \right)^3} + 9{\left( 0 \right)^2} - 2 = - 2$

Solution (4)

$f\left( x \right) = \sqrt {2x + 7} \\ f'\left( x \right) = \frac{1}{{2\sqrt {2x + 7} }}\left( 2 \right) = \frac{1}{{\sqrt {2x + 7} }}\\ f'\left( 1 \right) = \frac{1}{{\sqrt {2\left( 1 \right) + 7} }} = \frac{1}{{\sqrt 9 }} = \frac{1}{3}$

Solution (5)

$f\left( x \right) = 1 + \sqrt {{x^2} + 3x + 6} \\ f'\left( x \right) = 0 + \frac{1}{{2\sqrt {{x^2} + 3x + 6} }}\left( {2x + 3} \right) = \frac{{2x + 3}}{{2\sqrt {{x^2} + 3x + 6} }}\\ f'\left( 2 \right) = \frac{{2\left( 2 \right) + 3}}{{2\sqrt {{{\left( 2 \right)}^2} + 3\left( 2 \right) + 6} }} = \frac{7}{{2\sqrt {16} }} = \frac{7}{8}$

Solution (6)

$f\left( x \right) = \frac{x}{{x + 4}}\\ f'\left( x \right) = \frac{{\left( {x + 4} \right){{\left( x \right)}^\prime } - \left( x \right){{\left( {x + 4} \right)}^\prime }}}{{{{\left( {x + 4} \right)}^2}}}\\ f'\left( x \right) = \frac{{x + 4 - x}}{{{{\left( {x + 4} \right)}^2}}} = \frac{4}{{{{\left( {x + 4} \right)}^2}}}\\ f'\left( { - 3} \right) = \frac{4}{{{{\left( {\left( { - 3} \right) + 4} \right)}^2}}} = \frac{4}{{{1^2}}} = 4$

Solution (7)

$f\left( x \right) = \frac{{{x^2} + 3}}{{4 - {x^2}}}\\ f'\left( x \right) = \frac{{\left( {4 - {x^2}} \right)\left( {2x} \right) - \left( {{x^2} + 3} \right)\left( { - 2x} \right)}}{{{{\left( {4 - {x^2}} \right)}^2}}}\\ f'\left( x \right) = \frac{{8x - 2{x^3} + 2{x^3} + 6x}}{{{{\left( {4 - {x^2}} \right)}^2}}} = \frac{{14x}}{{{{\left( {4 - {x^2}} \right)}^2}}}\\ f'\left( { - 1} \right) = \frac{{14\left( { - 1} \right)}}{{{{\left( {4 - {{\left( { - 1} \right)}^2}} \right)}^2}}} = \frac{{ - 14}}{{{3^2}}} = - \frac{{14}}{9}$

Solution (8)

$f\left( x \right) = \sqrt x + \frac{3}{x} = \sqrt x + 3{x^{ - 1}}\\ f'\left( x \right) = \frac{1}{{2\sqrt x }} - 3{x^{ - 2}} = \frac{1}{{2\sqrt x }} - \frac{3}{{{x^2}}}\\ f'\left( 1 \right) = \frac{1}{{2\sqrt {\left( 1 \right)} }} - \frac{3}{{{{\left( 1 \right)}^2}}} = \frac{1}{2} - 3 = - \frac{5}{2}$

Solution (9)

$f\left( x \right) = 3 - \sqrt {5x} - \frac{9}{{2x}} = 3 - \sqrt {5x} - \frac{9}{2}{x^{ - 1}}\\ f'\left( x \right) = 0 - \frac{1}{{2\sqrt {5x} }}\left( 5 \right) + \frac{9}{2}{x^{ - 2}} = - \frac{{\sqrt 5 }}{{2\sqrt x }} + \frac{9}{{2{x^2}}}\\ f'\left( 2 \right) = - \frac{{\sqrt 5 }}{{2\sqrt {\left( 2 \right)} }} + \frac{9}{{2{{\left( 2 \right)}^2}}} = \frac{{\sqrt 5 }}{{2\sqrt 2 }} + \frac{9}{8}$

Solution (10)

$f\left( x \right) = 2{x^2} + 3x - 1\\ f'\left( x \right) = 4x + 3\\ f'\left( { - 1} \right) = 4\left( { - 1} \right) + 3 = - 1$