Using first principle in calculus.solve
2x²+5x+3
solve what? There is no equation.
But, since your subject is calculus, I assume you want the derivative.
If f(x) = 2x^2 + 5x + 3, then
f(x+h) = 2(x+h)^2 + 5(x+h) + 3
and df/dx is the limit as h→0 of
(2(x+h)^2 + 5(x+h) + 3 - (2x^2+5x+3))/h
= (4hx + 2h^2 + 5h)/h
= 4x + 5 + 2h
→ 4x+5
To find the derivative of the function f(x) = 2x²+5x+3 using the first principles of calculus, follow these steps:
Step 1: Begin with the definition of the derivative.
f'(x) = lim(h→0) [f(x+h) - f(x)] / h
Step 2: Substitute the given function into the definition.
f'(x) = lim(h→0) [(2(x+h)²+5(x+h)+3) - (2x²+5x+3)] / h
Step 3: Expand and simplify the expression.
f'(x) = lim(h→0) [(2x² + 4xh + 2h² + 5x + 5h + 3) - (2x² + 5x + 3)] / h
= lim(h→0) [2x² + 4xh + 2h² + 5x + 5h + 3 - 2x² - 5x - 3] / h
= lim(h→0) [4xh + 2h² + 5h] / h
Step 4: Cancel out the common factor of h.
f'(x) = lim(h→0) [h(4x + 2h + 5)] / h
Step 5: Simplify further.
f'(x) = lim(h→0) (4x + 2h + 5)
Step 6: Evaluate the limit as h approaches 0.
f'(x) = 4x + 5
Therefore, the derivative of the function f(x) = 2x²+5x+3 is f'(x) = 4x + 5.
To solve the given equation, 2x² + 5x + 3, using the first principle in calculus (also known as the limit definition of a derivative), follow these steps:
Step 1: Start by differentiating the equation using the first principle.
- The first principle states that the derivative of a function f(x) with respect to x is defined as:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
- Apply this definition to the given equation.
Step 2: Differentiate each term separately.
- Differentiate 2x²:
f'(x) = lim(h->0) [(2(x + h)² + 5(x + h) + 3) - (2x² + 5x + 3)] / h
Simplify:
f'(x) = lim(h->0) [(2x² + 4xh + 2h² + 5x + 5h + 3) - (2x² + 5x + 3)] / h
Remove parentheses:
f'(x) = lim(h->0) [2x² + 4xh + 2h² + 5x + 5h + 3 - 2x² - 5x - 3] / h
Simplify further:
f'(x) = lim(h->0) [4xh + 2h² + 5h] / h
Step 3: Simplify the expression:
f'(x) = lim(h->0) [h(4x + 2h + 5)] / h
Cancel out h:
f'(x) = lim(h->0) [4x + 2h + 5]
Step 4: Evaluate the limit as h approaches 0.
- Since h approaches 0, we can substitute h = 0 into the expression.
f'(x) = 4x + 2(0) + 5
Simplify:
f'(x) = 4x + 5
So, the derivative of the given equation, 2x² + 5x + 3, is 4x + 5.