Find the derivatives of the following functions using first principles: f(x) = x3 and f(x) = 1/x.

Question

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Here are the derivatives of the given functions using first principles:

1) $f(x) = x^3$

Write the definition of the derivative from first principles.
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substitute $f(x) = x^3$ and $f(x+h) = (x+h)^3$ into the formula.
$f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$
Expand $(x+h)^3$ and simplify the numerator.
$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ $f'(x) = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}$ $f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$
Factor out $h$ from the numerator and cancel it with the denominator.
$f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$ $f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)$
Apply the limit as $h \to 0$ .
$f'(x) = 3x^2 + 3x(0) + (0)^2$ $f'(x) = \boxed{\mathbf{3x^2}}$
2) $f(x) = \frac{1}{x}$
Write the definition of the derivative from first principles.
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substitute $f(x) = \frac{1}{x}$ and $f(x+h) = \frac{1}{x+h}$ into the formula.
$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$
Combine the fractions in the numerator.
$f'(x) = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h}$ $f'(x) = \lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}$
Simplify the complex fraction.
$f'(x) = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)}$ $f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)}$
Apply the limit as $h \to 0$ .
$f'(x) = \frac{-1}{x(x+0)}$ $f'(x) = \boxed{\mathbf{-\frac{1}{x^2}}}$
3) $f(x) = \sqrt{x}$
Write the definition of the derivative from first principles.
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substitute $f(x) = \sqrt{x}$ and $f(x+h) = \sqrt{x+h}$ into the formula.
$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$
Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+h} + \sqrt{x}$ .
$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$
Expand the numerator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ .
$f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$ $f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$
Cancel $h$ from the numerator and denominator.
$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$
Apply the limit as $h \to 0$ .
$f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}}$ $f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}}$ $f'(x) = \boxed{\mathbf{\frac{1}{2\sqrt{x}}}}$

Find the derivatives of the following functions using first principles: f(x) = x3 and f(x) = (1)/(x).

Step-by-step solution

Write the definition of the derivative from first principles.

Substitute $f(x) = x^3$ and $f(x+h) = (x+h)^3$ into the formula.

Expand $(x+h)^3$ and simplify the numerator.

Factor out $h$ from the numerator and cancel it with the denominator.

Apply the limit as $h \to 0$ .

Write the definition of the derivative from first principles.

Substitute $f(x) = \frac{1}{x}$ and $f(x+h) = \frac{1}{x+h}$ into the formula.

Combine the fractions in the numerator.

Simplify the complex fraction.

Apply the limit as $h \to 0$ .

Write the definition of the derivative from first principles.

Substitute $f(x) = \sqrt{x}$ and $f(x+h) = \sqrt{x+h}$ into the formula.

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+h} + \sqrt{x}$ .

Expand the numerator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ .

Cancel $h$ from the numerator and denominator.

Apply the limit as $h \to 0$ .

Still stuck on this one?

More Mathematics Questions

Find the derivatives of the following functions using first principles: f(x) = x3 and f(x) = (1)/(x).

Step-by-step solution

Write the definition of the derivative from first principles.

Substitute f(x)=x3f(x) = x^3f(x)=x3 and f(x+h)=(x+h)3f(x+h) = (x+h)^3f(x+h)=(x+h)3 into the formula.

Expand (x+h)3(x+h)^3(x+h)3 and simplify the numerator.

Factor out hhh from the numerator and cancel it with the denominator.

Apply the limit as h→0h \to 0h→0.

Write the definition of the derivative from first principles.

Substitute f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ and f(x+h)=1x+hf(x+h) = \frac{1}{x+h}f(x+h)=x+h1​ into the formula.

Combine the fractions in the numerator.

Simplify the complex fraction.

Apply the limit as h→0h \to 0h→0.

Write the definition of the derivative from first principles.

Substitute f(x)=xf(x) = \sqrt{x}f(x)=x​ and f(x+h)=x+hf(x+h) = \sqrt{x+h}f(x+h)=x+h​ into the formula.

Multiply the numerator and denominator by the conjugate of the numerator, which is x+h+x\sqrt{x+h} + \sqrt{x}x+h​+x​.

Expand the numerator using the difference of squares formula (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2(a−b)(a+b)=a2−b2.

Cancel hhh from the numerator and denominator.

Apply the limit as h→0h \to 0h→0.

Still stuck on this one?

More Mathematics Questions

Substitute $f(x) = x^3$ and $f(x+h) = (x+h)^3$ into the formula.

Expand $(x+h)^3$ and simplify the numerator.

Factor out $h$ from the numerator and cancel it with the denominator.

Apply the limit as $h \to 0$ .

Substitute $f(x) = \frac{1}{x}$ and $f(x+h) = \frac{1}{x+h}$ into the formula.

Apply the limit as $h \to 0$ .

Substitute $f(x) = \sqrt{x}$ and $f(x+h) = \sqrt{x+h}$ into the formula.

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+h} + \sqrt{x}$ .

Expand the numerator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ .

Cancel $h$ from the numerator and denominator.

Apply the limit as $h \to 0$ .