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)

1.

a) To find f'(a), we need to find the derivative of f(x) and then substitute x=a. Taking the derivative of f(x) = 4x^2 + 2x, we get f'(x) = 8x + 2. Substituting x=a, we have f'(a) = 8a + 2.

b) To determine the equation of the line tangent to the graph of f at (a,f(a)), we use the point-slope form of a line. The point (a,f(a)) is (1, 6) since a=1 and f(a) = 4(1)^2 + 2(1) = 6. The derivative is f'(x) = 8x + 2. Plugging in x=1, we get f'(1) = 8(1) + 2 = 10. Hence, the equation of the tangent line is y = 10(x - 1) + 6.

2.

To find f'(x) using the definition f'(x) = lim(h→0), we start by finding f(x + h). Given f(x) = √(x + 2), we have f(x + h) = √(x + h + 2). Now, we substitute f(x + h) and f(x) into the definition of f'(x):

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

To simplify the expression, we multiply the numerator and denominator by the conjugate:

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

Simplifying further, we get:

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

The term (x + h + 2) - (x + 2) simplifies to h. We can then cancel out the h in the numerator and denominator:

f'(x) = lim(h→0) [1 / (√(x + h + 2) + √(x + 2))].

Since h tends to 0 in the denominator, we can substitute h=0 into the expression:

f'(x) = 1 / (√(x + 2) + √(x + 2)).

Simplifying further, we get:

f'(x) = 1 / (2√(x + 2)).

Now, to determine the equation of the line tangent to the graph of f at (a, f(a)), we use the point-slope form of a line. The point (a, f(a)) is (2, √(2 + 2)) = (2, √4) = (2, 2) since a = 2 and f(a) = √(2 + 2) = √4 = 2. The derivative is f'(x) = 1 / (2√(x + 2)). Plugging in x=2, we get f'(2) = 1 / (2√(2 + 2)) = 1 / (2 * 2) = 1/4. Hence, the equation of the tangent line is y = (1/4)(x - 2) + 2.

3.

To find the derivative of g(r) = (9r^3 + 2r + 5)(r^2 + 6), we expand the expression using the distributive property:

g(r) = 9r^3(r^2 + 6) + 2r(r^2 + 6) + 5(r^2 + 6).

Simplifying further, we get:

g(r) = 9r^5 + 54r^3 + 2r^3 + 12r + 5r^2 + 30.

Now, to find the derivative, we take the derivative term by term:

g'(r) = d/dx(9r^5) + d/dx(54r^3) + d/dx(2r^3) + d/dx(12r) + d/dx(5r^2) + d/dx(30).

Simplifying each term, we get:

g'(r) = 45r^4 + 162r^2 + 6r^2 + 12 + 10r + 0.

Combining like terms, we obtain the final derivative:

g'(r) = 45r^4 + 168r^2 + 10r + 12.

1.

a) To find f'(a), we need to take the derivative of the function f(x) and evaluate it at x = a.
f(x) = 4x^2 + 2x
Taking the derivative using the power rule:
f'(x) = 8x + 2
Now substitute x = a:
f'(a) = 8a + 2

b) To find the equation of the line tangent to the graph of f at (a, f(a)), we need to use the point-slope form of a linear equation y - y1 = m(x - x1), where m represents the slope of the line.
The slope (m) is equal to f'(a) as found in part a.
So, m = f'(a) = 8a + 2
Using the given point (a, f(a)) = (1, f(1)), we substitute these values into the equation:
y - f(a) = (8a + 2)(x - a)

2.

To find f'(x) using the definition f'(x) = lim(h->0) [(f(x + h) - f(x)) / h], we need to find the limit of the difference quotient as h approaches 0.
f(x) = √(x + 2)
Now applying the definition of the derivative:
f'(x) = lim(h->0) [√(x + h + 2) - √(x + 2)] / h
To simplify this expression, we multiply the numerator and denominator by the conjugate:
f'(x) = lim(h->0) [(√(x + h + 2) - √(x + 2)) * (√(x + h + 2) + √(x + 2))] / (h * (√(x + h + 2) + √(x + 2)))
Expanding the numerator gives:
f'(x) = lim(h->0) (√(x + h + 2)^2 - √(x + 2)^2) / (h * (√(x + h + 2) + √(x + 2)))
Simplifying further:
f'(x) = lim(h->0) (x + h + 2 - x - 2) / (h * (√(x + h + 2) + √(x + 2)))
f'(x) = lim(h->0) (h) / (h * (√(x + h + 2) + √(x + 2)))
Canceling out the h:
f'(x) = lim(h->0) 1 / (√(x + h + 2) + √(x + 2))
Now substitute x = a:
f'(a) = 1 / (√(a + 2) + √(2))
Simplifying this further may require additional information about the value of a.

The equation of the line tangent to the graph of f at (a, f(a)) can be determined using the point-slope form of a linear equation as explained in question 1b.

3.

To find the derivative of the function g(r) = (9r^3 + 2r + 5)(r^2 + 6) by expanding the expression, we can use the product rule for derivatives.
The product rule states that if we have two functions u(r) and v(r), then the derivative of their product is given by the formula:
(uv)' = u'v + uv'

Applying the product rule to the given function:
g(r) = (9r^3 + 2r + 5)(r^2 + 6)
g'(r) = (9r^3 + 2r + 5)'(r^2 + 6) + (9r^3 + 2r + 5)(r^2 + 6)'

Now differentiate each term separately:
(9r^3 + 2r + 5)' = 27r^2 + 2
(r^2 + 6)' = 2r

Substituting the derivatives back into the equation:
g'(r) = (27r^2 + 2)(r^2 + 6) + (9r^3 + 2r + 5)(2r)

Expanding and simplifying further:
g'(r) = 27r^4 + 162r^2 + 2r^2 + 12 + 18r^4 + 4r^2 + 10r
g'(r) = 45r^4 + 166r^2 + 10r + 12

So, the derivative of g(r) is g'(r) = 45r^4 + 166r^2 + 10r + 12.

1.

a) To find f'(a), we need to find the derivative of the function f(x) and evaluate it at x = a. The derivative of f(x) = 4x^2 + 2x is given by f'(x) = 8x + 2.

To evaluate f'(a), substitute x = a into the derivative equation: f'(a) = 8a + 2.

b) We want to find the equation of the line tangent to the graph of f at the point (a, f(a)). The slope of the tangent line is equal to the derivative of f(x) at x = a, which is f'(a) = 8a + 2.

Since we have a point (a, f(a)), we can use the point-slope form of a line: y - f(a) = (8a + 2)(x - a).

Therefore, the equation of the tangent line is y - f(a) = (8a + 2)(x - a).

2.

To find f'(x) using the definition f'(x) = lim(h→0) [f(x + h) - f(x)] / h, we need to apply the limit to the difference quotient and simplify.

Given f(x) = √(x + 2), we can start by substituting x into the difference quotient: f'(x) = lim(h→0) [f(x + h) - f(x)] / h = lim(h→0) [√(x + h + 2) - √(x + 2)] / h.

Now, we need to simplify the difference quotient by using a common denominator: f'(x) = lim(h→0) [(√(x + h + 2) - √(x + 2)) * (√(x + h + 2) + √(x + 2))] / (h * (√(x + h + 2) + √(x + 2))).

Applying the difference of squares, we get: f'(x) = lim(h→0) [(x + h + 2) - (x + 2)] / (h * (√(x + h + 2) + √(x + 2))).

Simplifying further, we have: f'(x) = lim(h→0) (h) / (h * (√(x + h + 2) + √(x + 2))).

The h in the numerator and denominator cancels out: f'(x) = lim(h→0) 1 / (√(x + h + 2) + √(x + 2)).

Finally, we take the limit as h approaches 0: f'(x) = 1 / (√(x + 2) + √(x + 2)) = 1 / (2√(x + 2)).

To find the equation of the tangent line at (a, f(a)), we need to substitute x = a into the derivative, f'(x), and then apply the point-slope form of a line. Since a = 2, we have:

f'(x) = 1 / (2√(2 + 2)) = 1 / (2√4) = 1 / (2 * 2) = 1 / 4.

Using the point-slope form of a line, the equation of the tangent line is y - f(a) = f'(a)(x - a). Plugging in a = 2 and f'(a) = 1/4, we have:

y - f(2) = (1/4)(x - 2).

Simplifying further:

y - √(2 + 2) = 1/4(x - 2).

3.

To find the derivative of the function g(r) = (9r^3 + 2r + 5)(r^2 + 6), we need to expand the expression and then apply the power rule for derivatives.

Start by using the distributive property to expand the function:

g(r) = (9r^3 + 2r + 5)(r^2 + 6) = 9r^5 + 54r^3 + 2r^3 + 12r + 5r^2 + 30.

Now, apply the power rule for derivatives to each term individually:

The derivative of 9r^5 with respect to r is 45r^4.
The derivative of 54r^3 with respect to r is 162r^2.
The derivative of 2r^3 with respect to r is 6r^2.
The derivative of 12r with respect to r is 12.
The derivative of 5r^2 with respect to r is 10r.
The derivative of 30 (a constant term) with respect to r is 0.

Combining all the derivatives, we get:

g'(r) = 45r^4 + 162r^2 + 6r^2 + 12 + 10r + 0.

Simplifying further, we have:

g'(r) = 45r^4 + 168r^2 + 10r + 12.