Question 1

The equation of the line tangent to the graph of $f ( x ) = x ^ { 2 } + 3 x$ at x = 2 is

Accepted Answer

The slope of the tangent line to the graph of $f(x) = x^2 + 3x$ at any point x is given by the derivative $f'(x) = 2x + 3$. At $x = 2$, $f'(2) = 2(2) + 3 = 7$, so the slope is 7. The point on the graph at $x = 2$ is $f(2) = 2^2 + 3(2) = 4 + 6 = 10$, so we have the point (2, 10). The equation of the line in point-slope form is $y - y_1 = m(x - x_1)$, which becomes $y - 10 = 7(x - 2)$. Simplifying, $y = 7x - 14 + 10$, which gives $y = 7x - 4$.

Question 2

The equation of the line tangent to the graph of $f ( x ) = x ^ { 2 } + 8 x$ at x = 7 is

Accepted Answer

To find the equation of the tangent line at $x = 7$, first find the derivative of $f(x)$, which gives the slope of the tangent line. The derivative $f'(x) = 2x + 8$. Substituting $x = 7$ into $f'(x)$ gives $2(7) + 8 = 22$, so the slope is 22. Then, use the point-slope form of a line, $y - y_1 = m(x - x_1)$, with $m = 22$ and $x_1 = 7$, $y_1 = f(7) = 7^2 + 8(7) = 49 + 56 = 105$. Substituting these values gives $y - 105 = 22(x - 7)$, which simplifies to $y = 22x - 49$.

Question 3

The equation of the line tangent to the graph of $f ( x ) = 3 \sqrt { x }$ at x = 1 is

Accepted Answer

C

Question 4

For f (x) = 1 - x², find the slope of the secant line connecting the points whose x-coordinates are x = -4 and x = -3.9. Then use calculus to find the slope of the line that is tangent to the graph of f at x = -4.

Accepted Answer

The answer of For f (x) = 1 - x²,...

Question 5

For $f ( x ) = - \frac { 3 } { \sqrt { x } }$ , find the average rate of change of f (x) with respect to x as x changes from 144 to 145. Then use calculus to find the instantaneous rate of change at x = 144. Round your answer to six decimal places, if necessary.

Accepted Answer

The answer of For \(f ( x ) = -...

Question 6

If f (x) represents the price per barrel of oil in terms of time, what does $\frac { f \left( x _ { 0 } + h \right) - f \left( x _ { 0 } \right) } { h }$ represent? What about $\lim _ { h \rightarrow 0 } \frac { f \left( x _ { 0 } + h \right) - f \left( x _ { 0 } \right) } { h }$ ?

Accepted Answer

The answer of If f (x) represents the price per...

Question 7

Differentiating $f ( x ) = x ^ { 3 } - 3 x + 1$ gives $3 x ^ { 2 }$ .

Accepted Answer

Differentiating $f(x) = x^3 - 3x + 1$ gives $f'(x) = 3x^2 - 3$, not just $3x^2$.

Question 8

Differentiating $f ( x ) = x ^ { 6 } - 5 x + 2$ gives $6 x ^ { 5 }$ .

Accepted Answer

Differentiating $f(x) = x^6 - 5x + 2$ gives $6x^5 - 5$, because the derivative of $x^6$ is $6x^5$, the derivative of $-5x$ is $-5$, and the derivative of a constant (like $+2$) is $0$.

Question 9

Differentiate: $f ( x ) = x ^ { 8 } + 2$

Accepted Answer

The derivative of $x^8$ is $8x^7$ by applying the power rule, and the derivative of a constant (2) is 0, so the correct answer is $8x^7$.

Question 10

Differentiate: $f ( x ) = x ^ { 4 } + 7$

Accepted Answer

The derivative of $x^4$ is $4x^3$ by applying the power rule, and the derivative of a constant (7) is 0, so the correct answer is $4x^3$.

Question 11

Differentiating $f ( x ) = \frac { 1 } { 3 } x ^ { 7 } - 2 x ^ { 5 } + 9 x - 8$ gives $\frac { 7 x ^ { 6 } } { 3 } - 10 x ^ { 4 } + 9$ .

Accepted Answer

Differentiating $f(x) = \frac{1}{3}x^7 - 2x^5 + 9x - 8$ term by term using the power rule $d/dx[x^n] = nx^{n-1}$ gives $\frac{7}{3}x^6 - 10x^4 + 9$, which matches the provided expression.

Question 12

Differentiating $f ( x ) = \frac { 1 } { 5 } x ^ { 6 } - 3 x ^ { 4 } + 2 x - 4$ gives $\frac { 6 } { 5 } x ^ { 5 } - 12 x ^ { 3 } + 2$ .

Accepted Answer

Differentiating $f(x) = \frac{1}{5}x^6 - 3x^4 + 2x - 4$ term by term using the power rule $d/dx[x^n] = nx^{n-1}$ gives $\frac{6}{5}x^5 - 12x^3 + 2$, as the derivative of a constant term (-4) is 0.

Question 13

If $f ( x ) = \sqrt [ 3 ] { x } - \frac { 1 } { \sqrt { x } }$ , differentiate f (x).

Accepted Answer

The answer of If \(f ( x ) = \sqrt...

Question 14

Differentiate: $f ( x ) = \sqrt { x } + \frac { 1 } { \sqrt { x } }$

Accepted Answer

The derivative of $ \sqrt{x} $ is $ \frac{1}{2\sqrt{x}} $ and the derivative of $ \frac{1}{\sqrt{x}} $ is $ -\frac{1}{2\sqrt{x^3}} $.

Question 15

Differentiate: $f ( x ) = \sqrt { x } + \frac { 5 } { \sqrt { x } }$

Accepted Answer

The differentiation of $f(x) = \sqrt{x} + \frac{5}{\sqrt{x}}$ using the power rule gives $\frac{1}{2\sqrt{x}} - \frac{5}{2x^{3/2}}$, which simplifies to the expression in option A.

Question 16

Differentiate: $f ( x ) = \sqrt [ 7 ] { x } - \frac { 3 } { \sqrt { x } }$

Accepted Answer

The answer of Differentiate: \(f ( x ) = \sqrt...

Question 17

Differentiate: $f ( x ) = \frac { 2 } { 3 } x ^ { 6 } - \frac { 5 x } { 6 } + \frac { 2 } { 3 x } - \sqrt [ 5 ] { x }$

Accepted Answer

The answer of Differentiate: \(f ( x ) = \frac...

Question 18

Differentiate: $f ( x ) = \frac { 2 } { 3 } x ^ { 6 } - \frac { 7 } { 3 } x + \frac { 8 } { 7 x } - \sqrt [ 6 ] { x }$

Accepted Answer

The answer of Differentiate: \(f ( x ) = \frac...

Question 19

Find the equation of the tangent line to the curve $f ( x ) = x ^ { 3 } - x ^ { 2 } + 7$ at the point (1, 7).

Accepted Answer

The answer of Find the equation of the tangent line...

Question 20

Find the equation of the tangent line to the curve $f ( x ) = x ^ { 3 } - x ^ { 2 } + 1$ at the point (1, 1).

Accepted Answer

The answer of Find the equation of the tangent line...

Quiz 2: Differentiation: Basic Concepts

The equation of the line tangent to the graph of f(x) =x2+3xf ( x ) = x ^ { 2 } + 3 xf(x) =x2+3x at x = 2 is

The equation of the line tangent to the graph of f(x) =x2+8xf ( x ) = x ^ { 2 } + 8 xf(x) =x2+8x at x = 7 is

The equation of the line tangent to the graph of f(x) =3xf ( x ) = 3 \sqrt { x }f(x) =3x​ at x = 1 is

For f (x) = 1 - x2, find the slope of the secant line connecting the points whose x-coordinates are x = -4 and x = -3.9. Then use calculus to find the slope of the line that is tangent to the graph of f at x = -4.

For f(x) =−3xf ( x ) = - \frac { 3 } { \sqrt { x } }f(x) =−x​3​ , find the average rate of change of f (x) with respect to x as x changes from 144 to 145. Then use calculus to find the instantaneous rate of change at x = 144. Round your answer to six decimal places, if necessary.

Differentiating f(x)=x3−3x+1f ( x ) = x ^ { 3 } - 3 x + 1f(x)=x3−3x+1 gives 3x23 x ^ { 2 }3x2 .

Differentiating f(x)=x6−5x+2f ( x ) = x ^ { 6 } - 5 x + 2f(x)=x6−5x+2 gives 6x56 x ^ { 5 }6x5 .

Differentiate: f(x) =x8+2f ( x ) = x ^ { 8 } + 2f(x) =x8+2

Differentiate: f(x) =x4+7f ( x ) = x ^ { 4 } + 7f(x) =x4+7

Differentiating f(x)=13x7−2x5+9x−8f ( x ) = \frac { 1 } { 3 } x ^ { 7 } - 2 x ^ { 5 } + 9 x - 8f(x)=31​x7−2x5+9x−8 gives 7x63−10x4+9\frac { 7 x ^ { 6 } } { 3 } - 10 x ^ { 4 } + 937x6​−10x4+9 .

Differentiating f(x)=15x6−3x4+2x−4f ( x ) = \frac { 1 } { 5 } x ^ { 6 } - 3 x ^ { 4 } + 2 x - 4f(x)=51​x6−3x4+2x−4 gives 65x5−12x3+2\frac { 6 } { 5 } x ^ { 5 } - 12 x ^ { 3 } + 256​x5−12x3+2 .

If f(x)=x3−1xf ( x ) = \sqrt [ 3 ] { x } - \frac { 1 } { \sqrt { x } }f(x)=3x​−x​1​ , differentiate f (x).

Differentiate: f(x) =x+1xf ( x ) = \sqrt { x } + \frac { 1 } { \sqrt { x } }f(x) =x​+x​1​

Differentiate: f(x) =x+5xf ( x ) = \sqrt { x } + \frac { 5 } { \sqrt { x } }f(x) =x​+x​5​

Differentiate: f(x)=x7−3xf ( x ) = \sqrt [ 7 ] { x } - \frac { 3 } { \sqrt { x } }f(x)=7x​−x​3​

Differentiate: f(x)=23x6−5x6+23x−x5f ( x ) = \frac { 2 } { 3 } x ^ { 6 } - \frac { 5 x } { 6 } + \frac { 2 } { 3 x } - \sqrt [ 5 ] { x }f(x)=32​x6−65x​+3x2​−5x​

Differentiate: f(x)=23x6−73x+87x−x6f ( x ) = \frac { 2 } { 3 } x ^ { 6 } - \frac { 7 } { 3 } x + \frac { 8 } { 7 x } - \sqrt [ 6 ] { x }f(x)=32​x6−37​x+7x8​−6x​

Find the equation of the tangent line to the curve f(x)=x3−x2+7f ( x ) = x ^ { 3 } - x ^ { 2 } + 7f(x)=x3−x2+7 at the point (1, 7).

Find the equation of the tangent line to the curve f(x)=x3−x2+1f ( x ) = x ^ { 3 } - x ^ { 2 } + 1f(x)=x3−x2+1 at the point (1, 1).