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Math  /  Calculus

Question1. Use the definition to find the derivative of each of the following functions: (a) f(x):=x3f(x):=x^{3} for x∈Rx \in \mathbb{R}, (b) g(x):=1/xg(x):=1 / x for x∈R,x≠0x \in \mathbb{R}, x \neq 0, (c) h(x):=xh(x):=\sqrt{x} for x>0x>0, (d) k(x):=1/xk(x):=1 / \sqrt{x} for x>0x>0.

Studdy Solution

STEP 1

1. We are using the definition of the derivative, which is given by: $ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
2. We will apply this definition to each of the given functions.

STEP 2

1. Find the derivative of f(x)=x3 f(x) = x^3 .
2. Find the derivative of g(x)=1x g(x) = \frac{1}{x} .
3. Find the derivative of h(x)=x h(x) = \sqrt{x} .
4. Find the derivative of k(x)=1x k(x) = \frac{1}{\sqrt{x}} .

STEP 3

For f(x)=x3 f(x) = x^3 , use the definition of the derivative:
f′(x)=lim⁡h→0(x+h)3−x3h f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}

STEP 4

Expand (x+h)3 (x+h)^3 :
(x+h)3=x3+3x2h+3xh2+h3 (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
Substitute back into the derivative expression:
f′(x)=lim⁡h→0x3+3x2h+3xh2+h3−x3h f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}
f′(x)=lim⁡h→03x2h+3xh2+h3h f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}

STEP 5

Factor out h h from the numerator:
f′(x)=lim⁡h→0h(3x2+3xh+h2)h f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}
Cancel h h in the numerator and denominator:
f′(x)=lim⁡h→0(3x2+3xh+h2) f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)
Evaluate the limit as h→0 h \to 0 :
f′(x)=3x2 f'(x) = 3x^2

STEP 6

For g(x)=1x g(x) = \frac{1}{x} , use the definition of the derivative:
g′(x)=lim⁡h→01x+h−1xh g'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

STEP 7

Combine the fractions in the numerator:
g′(x)=lim⁡h→0x−(x+h)x(x+h)h g'(x) = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h}
g′(x)=lim⁡h→0−hhx(x+h) g'(x) = \lim_{h \to 0} \frac{-h}{hx(x+h)}
Cancel h h in the numerator and denominator:
g′(x)=lim⁡h→0−1x(x+h) g'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)}
Evaluate the limit as h→0 h \to 0 :
g′(x)=−1x2 g'(x) = -\frac{1}{x^2}

STEP 8

For h(x)=x h(x) = \sqrt{x} , use the definition of the derivative:
h′(x)=lim⁡h→0x+h−xh h'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}

STEP 9

Multiply the numerator and the denominator by the conjugate:
h′(x)=lim⁡h→0(x+h−x)(x+h+x)h(x+h+x) h'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}
h′(x)=lim⁡h→0(x+h)−xh(x+h+x) h'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}
h′(x)=lim⁡h→0hh(x+h+x) h'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}
Cancel h h in the numerator and denominator:
h′(x)=lim⁡h→01x+h+x h'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}
Evaluate the limit as h→0 h \to 0 :
h′(x)=12x h'(x) = \frac{1}{2\sqrt{x}}

STEP 10

For k(x)=1x k(x) = \frac{1}{\sqrt{x}} , use the definition of the derivative:
k′(x)=lim⁡h→01x+h−1xh k'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}

STEP 11

Combine the fractions in the numerator:
k′(x)=lim⁡h→0x−x+hhxx+h k'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}
Multiply the numerator and the denominator by the conjugate:
k′(x)=lim⁡h→0(x−x+h)(x+x+h)hxx+h(x+x+h) k'(x) = \lim_{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}
k′(x)=lim⁡h→0x−(x+h)hxx+h(x+x+h) k'(x) = \lim_{h \to 0} \frac{x - (x+h)}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}
k′(x)=lim⁡h→0−hhxx+h(x+x+h) k'(x) = \lim_{h \to 0} \frac{-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}
Cancel h h in the numerator and denominator:
k′(x)=lim⁡h→0−1xx+h(x+x+h) k'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}
Evaluate the limit as h→0 h \to 0 :
k′(x)=−12xx k'(x) = -\frac{1}{2x\sqrt{x}}
The derivatives of the functions are: (a) f′(x)=3x2 f'(x) = 3x^2 (b) g′(x)=−1x2 g'(x) = -\frac{1}{x^2} (c) h′(x)=12x h'(x) = \frac{1}{2\sqrt{x}} (d) k′(x)=−12xx k'(x) = -\frac{1}{2x\sqrt{x}}

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