Calculate the Derivatives at the Points Indicated:

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Exercise 1

f(x) =2x^2 - 6x + 5 at x = -5

Exercise 2

f(x) = x^3 + 2x - 5 at x = 1

Exercise 3

f(x) = \frac{1}{x} at x = 2

Exercise 4

f(x) = \sqrt{x} at x = 3

Exercise 5

f(x) = \frac{x}{x - 1} at x = 2.

Exercise 6

Given the equation f(x) = 2x² − 3x − 1, find the coordinates of the point on the curve in which the tangent forms a 45° angle with the x-axis.

 

 

Solution of exercise 1

Find the value of the derivative f(x) = 2x^2 - 6x + 5 at x = -5.

    \[f'(x) = \lim_ {h \rightarrow 0} \frac{2 (x + h)^2 - 6 (x + h) + 5 - (2x^2 - 6x + 5)}{h}\]

    \[= \lim_ {h \rightarrow 0} \frac{2x^2 + 4xh + 2h^2 - 6x - 6h + 5 - 2x^2 + 6x - 5}{h}\]

    \[= \lim_ {h \rightarrow 0} \frac{4xh - 6h + 2h^2}{h} = \lim_{h\rightarrow 0} \frac{h(4x - 6 + 2h)}{h} = 4x - 6\]

f'(-5) = 4 (-5) - 6 = -26

 

Solution of exercise 2

Find the value of the derivative f(x) = x^3 + 2x - 5  at x = 1.

    \[f' (x) = \lim_ {h \rightarrow 0} \frac{ (x + h)^3 + 2(x + h) - 5 - (x^3 + 2x - 5)} {h}\]

    \[=\lim_{h \rightarrow 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h - 5 - x^3 - 2x + 5}{h}\]

    \[=\lim_{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3 + 2h}{h}\]

    \[= \lim_{h \rightarrow 0} \frac{h(3x^2 + 3xh + h^2 + 2)}{h} = 3x^2 + 2\]

f'(1) = 3 (1)^2 + 2 = 5

 

Solution of exercise 3

Find the value of the derivative f(x) = \frac{1}{x} at x = 2.

    \[f'(x) = \lim_{h \rightarrow 0} \frac{\frac{1}{x + h} - \frac{1}{x}} {h}\]

    \[= \lim_{h \rightarrow 0} \frac{\frac{x - (x + h)}{x (x + h)}} {h}\]

    \[= \lim_ {h \rightarrow 0} \frac{\frac{-h}{x^2 + xh}} {h}\]

    \[= \lim_{h \rightarrow 0} (-\frac{1}{x^2 + xh}) = - \frac{1}{x^2}\]

f'(2) = -\frac{1}{2^2} = - \frac{1}{4}

 

Solution of exercise 4

Find the value of the derivative f(x) = \sqrt{x} at x = 3.

    \[f'(3) = \lim_{h \rightarrow 0} \frac{\sqrt{3 + h} - \sqrt{3}} {h}\]

    \[= \lim_{h \rightarrow 0} \frac{(\sqrt{3 + h }- \sqrt{3}) (\sqrt{3 + h} + \sqrt{3})} {h ( \sqrt{3 + h} + \sqrt{3})}\]

    \[= \lim_{h \rightarrow 0} \frac{(3 + h) - 3} {\sqrt{3 + h} + \sqrt{3}}\]

    \[= \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{3 + h} + \sqrt{3})}\]

    \[= \lim_{h \rightarrow 0} \frac{1}{\sqrt{3 + h} + \sqrt{3}} = \frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2 \sqrt{3}}\]

 

Solution of exercise 5

Find the value of the derivative f(x) = \frac{x}{x - 1} at x = 2.

    \[f'(x) = \lim_{h \rightarrow 0} \frac{\frac{x + h} {x + h - 1} - \frac{x}{x - 1}} {h}\]

    \[= \lim_{h \rightarrow 0} \frac{\frac{x^2 - x + hx - h - x^2 - hx + x}{(x + h - 1) (x - 1)}} {h}\]

    \[= \lim_ {h \rightarrow 0} \frac{\frac{-h}{(x + h - 1) (x - 1)}} {h} = \lim_{h \rightarrow 0} \frac{-1}{(x + h - 1) (x - 1)} = -\frac{1}{(x - 1)^2}\]

f'(2) = \frac{-1}{(2 - 1)^2} = -1

 

Solution of exercise 6

Given the equation f(x) = 2x² − 3x − 1, find the coordinates of the point on the curve in which the tangent forms a 45° angle with the x-axis .

f'(x) = tg 45^o = 1

    \[1 = \lim_{h \rightarrow 0} \frac{2 (x + h)^2 - 3(x + h) - 1 - (2x^2 - 3x - 1)} {h}\]

    \[1 = \lim_{h \rightarrow 0} \frac{2x^2 + 4xh + 2h^2 - 3x - 3h - 1- 2x^2 + 3x + 1} {h}\]

    \[1 = \lim_{h \rightarrow 0} \frac{h(4x + 2h - 3)}{h}\]

1 = 4x - 3         x = 1

f(1) = 2 \cdot 1^2 - 3 \cdot 1 - 1 = -2          P(1, -2)

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.