7.1 Determine $$ f^{\prime}(x) $$ from first principles if $$ f(x)=3 x^{2}-5 $$ Determine $$ \frac{d y}{d x} $$ if: $$ 7.2 .1 \quad y=2 x^{5}+\frac{4}{x^{3}} $$ 7.2.2 $$ \quad y=\left(\sqrt{x}-x^{2}\right)^{2} $$ QUESTION 8 Determine $$ f^{\prime}(x) $$ from first principles if $$ f(x)=-x^{2}+4 $$ Determine the derivative of: 8.2.1 $$ \quad y=3 x^{2}+10 x $$ 8.2 $$ \quad f(x)=\left(x-\frac{3}{2}\right. $$

Question

UpStudy AI · Accepted Answer

Simplify the expression by following steps:
- step0: Solution:
  $$1\times \left(x-\frac{3}{2}\right)+\left(x-\frac{3}{2}\right)\times 1$$
- step1: Multiply:
  $$x-\frac{3}{2}+\left(x-\frac{3}{2}\right)\times 1$$
- step2: Multiply:
  $$x-\frac{3}{2}+x-\frac{3}{2}$$
- step3: Add the terms:
  $$2x-3$$
Calculate or simplify the expression $$ (3*(x+h)^2 - 5 - (3*x^2 - 5))/h $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{\left(3\left(x+h\right)^{2}-5-\left(3x^{2}-5\right)\right)}{h}$$
- step1: Remove the parentheses:
  $$\frac{3\left(x+h\right)^{2}-5-\left(3x^{2}-5\right)}{h}$$
- step2: Subtract the terms:
  $$\frac{6xh+3h^{2}}{h}$$
- step3: Factor:
  $$\frac{h\left(6x+3h\right)}{h}$$
- step4: Reduce the fraction:
  $$6x+3h$$
Calculate or simplify the expression $$ 2*(\sqrt(x)-x^2)*(1/(2*\sqrt(x))-2*x) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$2\left(\sqrt{x}-x^{2}\right)\left(\frac{1}{\left(2\sqrt{x}\right)}-2x\right)$$
- step1: Remove the parentheses:
  $$2\left(\sqrt{x}-x^{2}\right)\left(\frac{1}{2\sqrt{x}}-2x\right)$$
- step2: Subtract the terms:
  $$2\left(\sqrt{x}-x^{2}\right)\times \frac{1-4x\sqrt{x}}{2\sqrt{x}}$$
- step3: Multiply the terms:
  $$\frac{1-4x\sqrt{x}}{\sqrt{x}}\times \left(\sqrt{x}-x^{2}\right)$$
- step4: Multiply the terms:
  $$\frac{\left(1-4x\sqrt{x}\right)\left(\sqrt{x}-x^{2}\right)}{\sqrt{x}}$$
- step5: Multiply the terms:
  $$\frac{\sqrt{x}-5x^{2}+4x^{3}\sqrt{x}}{\sqrt{x}}$$
- step6: Simplify:
  $$1-5x\sqrt{x}+4x^{3}$$
Calculate or simplify the expression $$ (-2*(x+h) + 0 - (-2*x + 4))/h $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{\left(-2\left(x+h\right)+0-\left(-2x+4\right)\right)}{h}$$
- step1: Remove the parentheses:
  $$\frac{-2\left(x+h\right)+0-\left(-2x+4\right)}{h}$$
- step2: Remove the parentheses:
  $$\frac{-2\left(x+h\right)+0+2x-4}{h}$$
- step3: Remove 0:
  $$\frac{-2\left(x+h\right)+2x-4}{h}$$
- step4: Calculate:
  $$\frac{-2h-4}{h}$$
- step5: Rewrite the fraction:
  $$-\frac{2h+4}{h}$$
Calculate or simplify the expression $$ 10*x^4 - 12/x^4 $$.
Simplify the expression by following steps:
- step0: Solution:
  $$10x^{4}-\frac{12}{x^{4}}$$
- step1: Reduce fractions to a common denominator:
  $$\frac{10x^{4}\times x^{4}}{x^{4}}-\frac{12}{x^{4}}$$
- step2: Transform the expression:
  $$\frac{10x^{4}\times x^{4}-12}{x^{4}}$$
- step3: Multiply the terms:
  $$\frac{10x^{8}-12}{x^{4}}$$
Calculate or simplify the expression $$ 6*x + 10 $$.
Simplify the expression by following steps:
- step0: Solution:
  $$6x+10$$
Let's solve the problems step by step.

### 7.1 Determine $$ f^{\prime}(x) $$ from first principles if $$ f(x)=3 x^{2}-5 $$

To find the derivative from first principles, we use the definition of the derivative:

$$
f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

1. Calculate $$ f(x+h) $$:
   $$
   f(x+h) = 3(x+h)^2 - 5 = 3(x^2 + 2xh + h^2) - 5 = 3x^2 + 6xh + 3h^2 - 5
   $$

2. Now, substitute into the derivative formula:
   $$
   f^{\prime}(x) = \lim_{h \to 0} \frac{(3x^2 + 6xh + 3h^2 - 5) - (3x^2 - 5)}{h}
   $$
   Simplifying this gives:
   $$
   f^{\prime}(x) = \lim_{h \to 0} \frac{6xh + 3h^2}{h} = \lim_{h \to 0} (6x + 3h)
   $$

3. Taking the limit as $$ h \to 0 $$:
   $$
   f^{\prime}(x) = 6x
   $$

### 7.2.1 Determine $$ \frac{d y}{d x} $$ if $$ y=2 x^{5}+\frac{4}{x^{3}} $$

Using the power rule and the quotient rule:
$$
\frac{d y}{d x} = \frac{d}{dx}(2x^5) + \frac{d}{dx}(4x^{-3})
$$
Calculating each term:
$$
\frac{d}{dx}(2x^5) = 10x^4
$$
$$
\frac{d}{dx}(4x^{-3}) = -12x^{-4}
$$
Combining these gives:
$$
\frac{d y}{d x} = 10x^4 - \frac{12}{x^4}
$$

### 7.2.2 Determine $$ \frac{d y}{d x} $$ if $$ y=\left(\sqrt{x}-x^{2}\right)^{2} $$

Using the chain rule:
Let $$ u = \sqrt{x} - x^2 $$, then $$ y = u^2 $$.
$$
\frac{d y}{d x} = 2u \cdot \frac{du}{dx}
$$
Calculating $$ \frac{du}{dx} $$:
$$
\frac{du}{dx} = \frac{1}{2\sqrt{x}} - 2x
$$
Thus,
$$
\frac{d y}{d x} = 2(\sqrt{x} - x^2)\left(\frac{1}{2\sqrt{x}} - 2x\right)
$$
Simplifying gives:
$$
\frac{d y}{d x} = 1 - 5x\sqrt{x} + 4x^3
$$

### 8. Determine $$ f^{\prime}(x) $$ from first principles if $$ f(x)=-x^{2}+4 $$

Using the same definition of the derivative:
1. Calculate $$ f(x+h) $$:
   $$
   f(x+h) = -(x+h)^2 + 4 = -x^2 - 2xh - h^2 + 4
   $$

2. Substitute into the derivative formula:
   $$
   f^{\prime}(x) = \lim_{h \to 0} \frac{(-x^2 - 2xh - h^2 + 4) - (-x^2 + 4)}{h}
   $$
   Simplifying gives:
   $$
   f^{\prime}(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h} = \lim_{h \to 0} (-2x - h)
   $$

3. Taking the limit as $$ h \to 0 $$:
   $$
   f^{\prime}(x) = -2x
   $$

### 8.2.1 Determine the derivative of $$ y=3 x^{2}+10 x $$

Using the power rule:
$$
\frac{d y}{d x} = 6x + 10
$$

### 8.2.2 Determine the derivative of $$ f(x)=\left(x-\frac{3}{2}\right) $$

Using the product rule:
$$
\frac{d y}{d x} = 1
$$

### Summary of Results
- $$ f^{\prime}(x) = 6x $$
- $$ \frac{d y}{d x} = 10x^4 - \frac{12}{x^4} $$
- $$ \frac{d y}{d x} = 1 - 5x\sqrt{x} + 4x^3 $$
- $$ f^{\prime}(x) = -2x $$
- $$ \frac{d y}{d x} = 6x + 10 $$
- $$ \frac{d y}{d x} = 1 $$
- $$ f^{\prime}(x) = 6x $$ - $$ \frac{d y}{d x} = 10x^4 - \frac{12}{x^4} $$ - $$ \frac{d y}{d x} = 1 - 5x\sqrt{x} + 4x^3 $$ - $$ f^{\prime}(x) = -2x $$ - $$ \frac{d y}{d x} = 6x + 10 $$ - $$ \frac{d y}{d x} = 1 $$

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions