Differentiation from first principles is a method used to find the derivative of a function by directly applying the definition of a derivative. This method is often used when the function is not easily differentiable using standard rules or techniques.

The general formula for finding the derivative of a function f(x) from first principles is:

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

**To apply this formula and find the derivative of a function, follow these steps:**

Step 1: Identify the function f(x) for which you want to find the derivative.

Step 2: Write down the general formula for differentiation from first principles.

Step 3: Substitute the function f(x) into the formula.

Step 4: Simplify the expression by expanding and combining like terms.

Step 5: Take the limit as h approaches 0 by substituting h = 0 into the expression.

Step 6: Evaluate the limit to find the derivative of the function.

Step 7: Verify the result by comparing it with known derivative rules or using other methods to find the derivative.

By following these steps, you can find the derivative of a function using the differentiation from first principles method. This method provides a rigorous way to find the derivative of a function by directly applying the definition of a derivative.

**Question 1:**

Find the derivative of f(x) = 3x^2 - 2x + 5 from first principles.

**Solution 1:**

Given function: f(x) = 3x^2 - 2x + 5

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

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

Substitute f(x) into the formula:

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

Expand and simplify the expression:

f'(x) = \lim_{{h \to 0}} \frac{3(x^2 + 2xh + h^2) - 2x - 2h + 5 - 3x^2 + 2x - 5}{h}

f'(x) = \lim_{{h \to 0}} \frac{3x^2 + 6xh + 3h^2 - 2x - 2h + 5 - 3x^2 + 2x - 5}{h}

f'(x) = \lim_{{h \to 0}} \frac{6xh + 3h^2 - 2h}{h}

f'(x) = \lim_{{h \to 0}} 6x + 3h - 2

f'(x) = 6x - 2

Therefore, the derivative of f(x) = 3x^2 - 2x + 5 is f'(x) = 6x - 2.

**Question 2:**

Find the derivative of g(x) = 4x^3 + 2x^2 - x from first principles.

**Solution 2:**

Given function: g(x) = 4x^3 + 2x^2 - x

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

g'(x) = \lim_{{h \to 0}} \frac{g(x+h) - g(x)}{h}

Substitute g(x) into the formula:

g'(x) = \lim_{{h \to 0}} \frac{4(x+h)^3 + 2(x+h)^2 - (x+h) - (4x^3 + 2x^2 - x)}{h}

Expand and simplify the expression:

g'(x) = \lim_{{h \to 0}} \frac{4(x^3 + 3x^2h + 3xh^2 + h^3) + 2(x^2 + 2xh + h^2) - x - h - 4x^3 - 2x^2 + x}{h}

g'(x) = \lim_{{h \to 0}} \frac{4x^3 + 12x^2h + 12xh^2 + 4h^3 + 2x^2 + 4xh + 2h^2 - x - h - 4x^3 - 2x^2 + x}{h}

g'(x) = \lim_{{h \to 0}} \frac{12x^2h + 12xh^2 + 4h^3 + 4xh + 2h^2 - h}{h}

g'(x) = \lim_{{h \to 0}} 12x^2 + 12xh + 4h^2 + 4x + 2h - 1

g'(x) = 12x^2 + 4x - 1

Therefore, the derivative of g(x) = 4x^3 + 2x^2 - x is

g'(x) = 12x^2 + 4x - 1

**Question 3:**

Find the derivative of h(x) = sin(x) from first principles.

**Solution 3:**

Given function: h(x) = \sin(x)

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

h'(x) = \lim_{{h \to 0}} \frac{h(x+h) - h(x)}{h}

Substitute h(x) into the formula:

h'(x) = \lim_{{h \to 0}} \frac{\sin(x+h) - \sin(x)}{h}

Apply the angle sum identity for sine function:

h'(x) = \lim_{{h \to 0}} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}

Simplify the expression:

h'(x) = \lim_{{h \to 0}} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}

h'(x) = \lim_{{h \to 0}} \frac{\sin(x)(\cos(h) - 1) + \cos(x)\sin(h)}{h}

h'(x) = \lim_{{h \to 0}} \frac{\sin(x)(\cos(h) - 1)}{h} + \lim_{{h \to 0}} \frac{\cos(x)\sin(h)}{h}

h'(x) = \sin(x)\lim_{{h \to 0}} \frac{\cos(h) - 1}{h} + \cos(x)\lim_{{h \to 0}} \frac{\sin(h)}{h}

Using the limit definition of derivative for sine function, we know:

\lim_{{h \to 0}} \frac{\sin(h)}{h} = 1

Therefore, the derivative of h(x) = sin(x) is h'(x) = cos(x).

**Question 4:**

Find the derivative of k(x) = \sqrt{x} from first principles.

**Solution 4:**

Given function: k(x) = \sqrt{x}

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

k'(x) = \lim_{{h \to 0}} \frac{k(x+h) - k(x)}{h}

Substitute k(x) into the formula:

k'(x) = \lim_{{h \to 0}} \frac{\sqrt{x+h} - \sqrt{x}}{h}

Rationalize the numerator:

k'(x) = \lim_{{h \to 0}} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}

k'(x) = \lim_{{h \to 0}} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}

k'(x) = \lim_{{h \to 0}} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}

k'(x) = \lim_{{h \to 0}} \frac{1}{\sqrt{x+h} + \sqrt{x}}

k'(x) = \frac{1}{2\sqrt{x}}

Therefore, the derivative of k(x) = âˆšx is k'(x) = 1/(2âˆšx).

**Question 5:**

Find the derivative of y(x) = e^xfrom first principles.

**Solution 5:**

Given function: y(x) = e^x

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

y'(x) = \lim_{{h \to 0}} \frac{y(x+h) - y(x)}{h}

Substitute y(x) into the formula:

y'(x) = \lim_{{h \to 0}} \frac{e^{x+h} - e^x}{h}

Apply the properties of exponential functions:

y'(x) = \lim_{{h \to 0}} \frac{e^x \cdot e^h - e^x}{h}

y'(x) = \lim_{{h \to 0}} \frac{e^x (e^h - 1)}{h}

Since \lim_{{h \to 0}} \frac{e^h - 1}{h} = 1,

y'(x) = e^x

Therefore, the derivative of y(x) = e^x is y'(x) = e^x.

**Questions**

1. \frac{d}{dx}(f(x)g(x))

2. \frac{d}{dx}(3x^2 \cdot \sin(x))

3. \frac{d}{dx}(e^x \cdot \cos(x))

4. \frac{d}{dx}(2x \cdot \ln(x))

5. \frac{d}{dx}(x^3 \cdot e^{2x})

6. \frac{d}{dx}(4x \cdot \tan(x))

7. \frac{d}{dx}(6x^2 \cdot \sqrt{x})

8. \frac{d}{dx}(\ln(x) \cdot e^x)

9. \frac{d}{dx}(2x^3 \cdot \sin(2x))

10. \frac{d}{dx}(x \cdot \cos(x))

11. \frac{d}{dx}(4x^2 \cdot e^x)

12. \frac{d}{dx}(e^{3x} \cdot \tan(x))

13. \frac{d}{dx}(5x \cdot \ln(2x))

14. \frac{d}{dx}(x^4 \cdot \cosh(x))

15. \frac{d}{dx}(e^{-x} \cdot \sin(x))

**Answer:**

1. g(x)+f(x)g'(x)

2. 6x\sin(x)+3x^2\cos(x)

3. e^x\cos(x)-e^x\sin(x)

4. 2\ln(x)+2

5. 3x^2e^{2x}+2x^3e^{2x}

6. 4\tan(x)+4x\sec^2(x)

7. 12x\sqrt{x}+3x^2\sqrt{x}

8. \frac{1}{x}e^x+\ln(x)e^x

9. 6x^2\sin(2x)+4x^3\cos(2x)

10. \cos(x)-x\sin(x)

11. 4x^2e^x+8xe^x

12. 3e^{3x}\tan(x)+e^{3x}\sec^2(x)

13. 5\ln(2x)+5 [\frac{1}{x}]

14. 4x^3\cosh(x)+x^4\sinh(x)

15. -e^{-x}\sin(x)-e^{-x}\cos(x)

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