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Use a derivative of function calculator with limit.

a) Find the derivative of f(x) = 3x^3 using the limit
b) Determine the slope of the tangent line to g(x) = e^x at x = 1
c) Calculate the instantaneous rate of change of h(x) = ln(x) at x = 2
d) Find the derivative of k(x) = cos(x) using the limit definition

1 Answer

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Final Answers:

a)
\(f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h) = 9x^2\)

b) The slope of the tangent line to
\(g(x) = e^x\) at \(x = 1\) is \(e\) .

c) The instantaneous rate of change of
\(h(x) = \ln(x)\) at \(x = 2\) is \(1/2\) .


d) \(k'(x) = \lim_(h \to 0) (k(x + h) - k(x))/(h) = -\sin(x)\)

Step-by-step explanation:

a) To find the derivative of
\(f(x) = 3x^3\) using the limit definition, we apply

the formula for the derivative:
\(f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)\) . For
\(f(x) = 3x^3\) ,

applying the limit definition gives us
\(f'(x) = \lim_(h \to 0) (3(x + h)^3 - 3x^3)/(h)\) ,

simplifying this yields
\(f'(x) = 9x^2\) .

b) The slope of the tangent line to
\(g(x) = e^x\) at
\(x = 1\) can be found by

evaluating the derivative of
\(g(x)\) at that point. The derivative of
\(e^x\) is

itself, so
\(g'(x) = e^x\) . At
\(x = 1\), \(g'(1) = e\) , indicating the slope of the

tangent line.

c) To determine the instantaneous rate of change of
\(h(x) = \ln(x)\) at


\(x = 2\) , we find its derivative. The derivative of
\(\ln(x)\) is
\(1/x\) . So, at
\(x = 2\) ,

the derivative
\(h'(2) = 1/2\) represents the instantaneous rate of change.

d) For
\(k(x) = \cos(x)\) , using the limit definition of the derivative,
\(k'(x) = \lim_(h \to 0) (\cos(x + h) - \cos(x))/(h)\) . Simplifying this limit

expression results in
\(k'(x) = -\sin(x)\) , giving us the derivative of
\(\cos(x)\).

User Kenhowardpdx
by
7.3k points

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