For the first function, y = 1/2x⁴, we will differentiate from the first principle.

Question

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Here are the derivatives of the given functions.

For the first function, $y = \frac{1}{2}x^4$ , we will differentiate from the first principle.
The first principle of differentiation is given by:

\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Identify $f(x)$ and $f(x+h)$ .
Given $f(x) = \frac{1}{2}x^4$ .
Then $f(x+h) = \frac{1}{2}(x+h)^4$ .
Expand $(x+h)^4$ .
$(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$
Substitute the expansion into $f(x+h)$ .
$f(x+h) = \frac{1}{2}(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4)$ $f(x+h) = \frac{1}{2}x^4 + 2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4$
Calculate $f(x+h) - f(x)$ .
$f(x+h) - f(x) = \left(\frac{1}{2}x^4 + 2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4\right) - \frac{1}{2}x^4$ $f(x+h) - f(x) = 2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4$
Divide by $h$ .
$\frac{f(x+h) - f(x)}{h} = \frac{2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4}{h}$ $\frac{f(x+h) - f(x)}{h} = 2x^3 + 3x^2h + 2xh^2 + \frac{1}{2}h^3$
Take the limit as $h \to 0$ .
$\frac{dy}{dx} = \lim_{h \to 0} (2x^3 + 3x^2h + 2xh^2 + \frac{1}{2}h^3)$ $\frac{dy}{dx} = 2x^3 + 3x^2(0) + 2x(0)^2 + \frac{1}{2}(0)^3$ $\frac{dy}{dx} = 2x^3$
The derivative of the first function is:
$\boxed{\frac{dy}{dx} = 2x^3}$
For the second function, $y = \sqrt{x}(1+2x)^2$ :
Rewrite the function using exponent notation.
$y = x^{1/2}(1+2x)^2$
Apply the product rule, $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ .
Let $u = x^{1/2}$ and $v = (1+2x)^2$ .
Find $\frac{du}{dx}$ .
$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
Find $\frac{dv}{dx}$ using the chain rule.
Let $w = 1+2x$ , so $v = w^2$ .
$\frac{dv}{dw} = 2w$ $\frac{dw}{dx} = \frac{d}{dx}(1+2x) = 2$ $\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx} = 2w \cdot 2 = 4w = 4(1+2x)$
Substitute $u, v, \frac{du}{dx}, \frac{dv}{dx}$ into the product rule formula.
$\frac{dy}{dx} = x^{1/2} \cdot 4(1+2x) + (1+2x)^2 \cdot \frac{1}{2}x^{-1/2}$ $\frac{dy}{dx} = 4\sqrt{x}(1+2x) + \frac{(1+2x)^2}{2\sqrt{x}}$
Simplify the expression by finding a common denominator.
$\frac{dy}{dx} = \frac{4\sqrt{x}(1+2x) \cdot 2\sqrt{x}}{2\sqrt{x}} + \frac{(1+2x)^2}{2\sqrt{x}}$ $\frac{dy}{dx} = \frac{8x(1+2x) + (1+2x)^2}{2\sqrt{x}}$
Factor out $(1+2x)$ from the numerator.
$\frac{dy}{dx} = \frac{(1+2x)[8x + (1+2x)]}{2\sqrt{x}}$ $\frac{dy}{dx} = \frac{(1+2x)(8x + 1 + 2x)}{2\sqrt{x}}$ $\frac{dy}{dx} = \frac{(1+2x)(10x + 1)}{2\sqrt{x}}$
The derivative of the second function is:
$\boxed{\frac{dy}{dx} = \frac{(1+2x)(10x + 1)}{2\sqrt{x}}}$

For the first function, y = (1)/(2)x⁴, we will differentiate from the first principle.

Answer

Step-by-step solution

Identify $f(x)$ and $f(x+h)$ .

Expand $(x+h)^4$ .

Substitute the expansion into $f(x+h)$ .

Calculate $f(x+h) - f(x)$ .

Divide by $h$ .

Take the limit as $h \to 0$ .

Rewrite the function using exponent notation.

Apply the product rule, $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ .

Find $\frac{du}{dx}$ .

Find $\frac{dv}{dx}$ using the chain rule.

Substitute $u, v, \frac{du}{dx}, \frac{dv}{dx}$ into the product rule formula.

Simplify the expression by finding a common denominator.

Factor out $(1+2x)$ from the numerator.

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More Computer Science Questions

For the first function, y = (1)/(2)x⁴, we will differentiate from the first principle.

Answer

Step-by-step solution

Identify f(x)f(x)f(x) and f(x+h)f(x+h)f(x+h).

Expand (x+h)4(x+h)^4(x+h)4.

Substitute the expansion into f(x+h)f(x+h)f(x+h).

Calculate f(x+h)−f(x)f(x+h) - f(x)f(x+h)−f(x).

Divide by hhh.

Take the limit as h→0h \to 0h→0.