Please I am lost on how to find the derivative of function using first principal,could someone please work out an example on how to solve such problem?
Example:
f(x)=x²
f(x+dx)
=(x+dx)³
=x³+3x²dx+3x(dx)²+(dx)³
[f(x+dx)-f(x)]/dx
=[x³+3x²dx+3x(dx)²+(dx)³ - x³]/dx
=[3x²dx+3x(dx)²+(dx)³]/dx
=3x²+3x(dx)+(dx)²
Take limit when dx->0
f'(x)
=[f(x+dx)-f(x)]/dx
=3x²
Note: For simplicity, above used dx instead of δx.
δx is meant to be finite, and
dx is generally taken to be infinitely small (after taking limit of δx->0)
correction:
problem is f(x)=x³
Of course! I'll walk you through an example of finding the derivative of a function using the first principles. Let's say we have the function f(x) = 2x^2 - 3x + 5.
Step 1: Write down the given function: f(x) = 2x^2 - 3x + 5.
Step 2: Write down the definition of the derivative using the first principles:
f'(x) = lim(h -> 0) [f(x + h) - f(x)] / h
Step 3: Substitute the given function into the definition of the derivative:
f'(x) = lim(h -> 0) [2(x + h)^2 - 3(x + h) + 5 - (2x^2 - 3x + 5)] / h
Step 4: Simplify the expression inside the limit:
f'(x) = lim(h -> 0) [2(x^2 + 2hx + h^2) - 3x - 3h + 5 - 2x^2 + 3x - 5] / h
Step 5: Expand and simplify the expression inside the limit:
f'(x) = lim(h -> 0) [2x^2 + 4hx + 2h^2 - 3x - 3h + 5 - 2x^2 + 3x - 5] / h
f'(x) = lim(h -> 0) [4hx + 2h^2 - 3h] / h
Step 6: Cancel out common terms to simplify further:
f'(x) = lim(h -> 0) [h(4x + 2h - 3)] / h
Step 7: Eliminate the h term in the numerator by canceling it out:
f'(x) = lim(h -> 0) 4x + 2h - 3
Step 8: Take the limit as h approaches 0:
f'(x) = 4x - 3
Congratulations! You have found the derivative of the given function f(x) = 2x^2 - 3x + 5 using the first principles.