Question

using the first principles, show that the derivative of f(x)=mx+b is m

Answer 1

To find the derivative of a function using first principles, we use the definition:

f'(x) = lim(h→0) [ f(x + h) - f(x) ] / h

Let's apply this definition to the function f(x) = mx + b:

f'(x) = lim(h→0) [ f(x + h) - f(x) ] / h

Now substitute f(x) = mx + b:

f'(x) = lim(h→0) [ (m(x + h) + b) - (mx + b) ] / h

Simplify the expression inside the limit:

f'(x) = lim(h→0) [ mx + mh + b - mx - b ] / h

Combine like terms:

f'(x) = lim(h→0) [ mh ] / h

Cancel out the h:

f'(x) = lim(h→0) m

Since h approaches 0, the limit is just m:

f'(x) = m

Therefore, the derivative of f(x) = mx + b with respect to x is m.