Final answer:
a) The derivative of f(x) = x^2 is f'(x) = 2x. b) The derivative of g(x) = sin(x) is g'(x) = cos(x). c) The slope of the tangent line to h(x) = √x at x = 4 is 1/4. d) The instantaneous rate of change of k(x) = 2x^2 at x = 3 is 12.
Step-by-step explanation:
a) Derivative of f(x) = x^2 using the limit definition:
Using the limit definition of a derivative, we have:
f'(x) = lim(h->0) [(f(x+h) - f(x)) / h]
Substituting f(x) = x^2, we get:
f'(x) = lim(h->0) [((x+h)^2 - x^2) / h]
Simplifying, we have:
f'(x) = lim(h->0) [(2xh + h^2) / h]
= lim(h->0) [2x + h]
= 2x
Therefore, the derivative of f(x) = x^2 is f'(x) = 2x.
b) Derivative of g(x) = sin(x) using the limit definition:
Using the limit definition, we have:
g'(x) = lim(h->0) [(g(x+h) - g(x)) / h]
Substituting g(x) = sin(x), we get:
g'(x) = lim(h->0) [(sin(x+h) - sin(x)) / h]
Using the trigonometric identities, we can simplify further to get:
g'(x) = cos(x)
Therefore, the derivative of g(x) = sin(x) is g'(x) = cos(x).
c) Slope of the tangent line to h(x) = √x at x = 4:
To find the slope of the tangent line, we need to find the derivative of h(x) = √x and evaluate it at x = 4. Using the power rule for derivatives, the derivative of h(x) = √x is:
h'(x) = (1/2) * x^(-1/2)
Evaluating h'(x) at x = 4:
h'(4) = (1/2) * 4^(-1/2)
= (1/2) * (1/√4)
= 1/4
Therefore, the slope of the tangent line to h(x) = √x at x = 4 is 1/4.
d) Instantaneous rate of change of k(x) = 2x^2 at x = 3:
The instantaneous rate of change is the same as the derivative of the function. Taking the derivative of k(x) = 2x^2, we get:
k'(x) = 4x
Evaluating k'(x) at x = 3:
k'(3) = 4 * 3
= 12
Therefore, the instantaneous rate of change of k(x) = 2x^2 at x = 3 is 12.