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) = √x is k'(x) = 1/(2√x).
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)
