1.

a) For the Function and point below , Find f’(a).

b) Determine the equation of the line tangent to the graph of f at (a,f(a)) for the given value of

f(x) = 4x2+2x, a =1

F’(a) =

y =

2.

For the function find f’ using the definition f’(x) = lim(h→0)

Determine the equation of the line tangent to the graph of f at (a,f(a)) for the given value of

F(x) = √x+2, a=2

f’(x) =

y =

3.

find the derivative of the following function by expanding the expression

g(r) = (9r3+2r+5)(r2+6)

1a) To find f'(a), we need to take the derivative of the given function f(x) = 4x^2 + 2x and evaluate it at x = a.

The derivative of f(x) is given by f'(x) = 8x + 2.

Hence, f'(a) = 8a + 2.

1b) To determine the equation of the line tangent to the graph of f at (a, f(a)), we can use the point-slope form of a line.

The slope of the tangent line is equal to the derivative of f at the point (a, f(a)), which we found to be f'(a) = 8a + 2.

So, the equation of the tangent line is y - f(a) = f'(a) * (x - a).

Substituting the values f(a) = 4a^2 + 2a and f'(a) = 8a + 2, we get:

y - (4a^2 + 2a) = (8a + 2) * (x - a).

Simplifying this equation will give us the final equation of the tangent line.

2a) To find f'(x) using the definition f'(x) = lim(h→0) [f(x + h) - f(x)] / h, we substitute the given function f(x) = √x + 2 into the definition.

We have:

f'(x) = lim(h→0) [(√(x + h) + 2) - (√x + 2)] / h.

Solving this limit will give us f'(x).

2b) To find the equation of the tangent line to the graph of f at (a, f(a)), we will first evaluate f'(x) at x = a to get f'(a).

Then, we can use the point-slope form of a line to determine the equation of the tangent line.

The equation will be y - f(a) = f'(a) * (x - a).

Substituting the values f(a) and f'(a), we can simplify the equation to get the final result.

3) To find the derivative of the function g(r) = (9r^3 + 2r + 5)(r^2 + 6), we can use the product rule.

The product rule states that if f(x) = g(x) * h(x), then f'(x) = g'(x) * h(x) + g(x) * h'(x).

For this problem, let's find the derivatives of (9r^3 + 2r + 5) and (r^2 + 6) individually.

The derivative of (9r^3 + 2r + 5) with respect to r will give us g'(r).

The derivative of (r^2 + 6) with respect to r will give us h'(r).

Then, we can use the product rule formula to calculate g'(r) * (r^2 + 6) + (9r^3 + 2r + 5) * h'(r).

Simplifying this expression will give us the derivative of g(r).

where do you get stuck? By the time you get to calculus, doing something like expanding a product of polynomials or evaluating f(1) should be no problem.