Exercises
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Exercises
Find the derivatives of the following functions.
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The second derivative of a function, written $f''(x)$, is the derivative of the first derivative. Find the second derivative of the following functions.
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Explorations
- Show that the graph of $f(x) = x^3 + x^2 + x + 1$ has no horizontal tangent lines.
- Find values of a, b and c such that $f(x) = ax^2 + bx + c$ is tangent to $y = 4x - 5$ at $x = 0$ and $y=-4x - 13$ at $x=-2$.
- Find values of a, b and c such that $f(x) = ax^3 + bx + c$ is tangent to $y = 13$ at $x = -1$ and $y=13x - 7$ at $x=1$.
The normal line to a curve at a point is the line that's perpendicular to the tangent line at that point. Find the equation of the normal line to each function at the given point. Remember that two lines are perpendicular if and only if their slopes are negative reciprocals of each other.
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Find the derivative of $f(x) = 4x^3 - 3x^2 + 2x - 1$.
$$f'(x) = \frac{d}{dx}4x^3 - \frac{d}{dx}3x^2 + \frac{d}{dx}2x - \frac{d}{dx}1$$ $$f'(x) = 4\frac{d}{dx}x^3 - 3\frac{d}{dx}x^2 + 2\frac{d}{dx}x - \frac{d}{dx}1$$ $$f'(x) = 4 \cdot 3x^2 - 3 \cdot 2x + 2 \cdot 1 - 0$$ $$f'(x) = 12x^2 - 6x + 2$$
Find the derivative of $f(x) = \sqrt{x} + \sqrt[3]{x}$
$$f'(x) = \frac{d}{dx}\sqrt{x} + \frac{d}{dx}\sqrt[3]{x}$$ $$f'(x) = \frac{d}{dx}x^{1/2} + \frac{d}{dx}x^{3/2}$$ $$f'(x) = \frac{1}{2}x^{-1/2} + \frac{3}{2}x^{1/2}$$ $$f'(x) = \frac{1}{2}x^{-1/2} + \frac{3}{2}x^{1/2}$$
Find the derivative of $f(x) = \cos(x) + \sin(x)$
$$f'(x) = -\sin(x) + \cos(x)$$
Find the derivative of $f(x) = x^{-1/2}$.
$$f'(x) = -\frac{1}{2}x^{-3/2}$$
Find the derivative of $\frac{x + \sqrt{x}}{x}$.
Split the numerator up into two fractions first then simplify the one with the square root by writing it using a fractional exponent.
Find the derivative of $\frac{x + \sqrt{x}}{x}$.
$$f'(x) = \frac{d}{dx}\left(\frac{x + \sqrt{x}}{x}\right)$$ $$f'(x) = \frac{d}{dx}\left(1 + \frac{\sqrt{x}}{x}\right)$$ $$f'(x) = \frac{d}{dx}\left(1 + x^{-1/2}\right)$$ $$f'(x) = \frac{d}{dx}1 + \frac{d}{dx}x^{-1/2}$$ $$f'(x) = 0 + \frac{-1}{2}x^{-3/2}$$ $$f'(x) = -\frac{1}{2}x^{-3/2}$$
Find the derivative of $g(x) = \sqrt{x}(x - 1)$.
$$\begin{aligned} f'(x) &= \frac{d}{dx}\left(\sqrt{x}(x - 1)\right) \\ &= \frac{d}{dx}\left(x^{1/2}(x - 1)\right) \\ &= \frac{d}{dx}\left(x^{3/2} - x^{1/2}\right) \\ &= \frac{d}{dx}x^{3/2} - \frac{d}{dx}x^{1/2} \\ &= \frac{3}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \\ &= \frac{1}{2}x^{-1/2}\left(3x - 1\right)\\ &= \frac{3x - 1}{2x^{1/2}} \\ &= \frac{3x - 1}{2\sqrt{x}} \\ &= \frac{3x\sqrt{x} - \sqrt{x}}{2x} \end{aligned}$$
In my class, you would be forgiven for stopping at the second to the last line and not rationalizing the denominator. For practical purposes, that line will be more useful for the things we're going to be doing later in the class.
Find the second derivative of $f(x) = x^2 + 2$.
$$f'(x) = 2x$$ $$f''(x) = \frac{d}{dx}f'(x)$$ $$f''(x) = \frac{d}{dx}(2x)$$ $$f''(x) = 2$$
Find the second derivative of $f(x) = 2x - 1$.
$$f'(x) = 2$$ $$f''(x) = \frac{d}{dx}f'(x)$$ $$f''(x) = \frac{d}{dx}(2)$$ $$f''(x) = 0$$
Find the second derivative of $f(x) = \cos(x) - 2\sin(x)$.
$$f'(x) = -\sin(x) - 2\cos(x)$$ $$f''(x) = \frac{d}{dx}f'(x)$$ $$f''(x) = \frac{d}{dx}(-\sin(x) - 2\cos(x))$$ $$f''(x) = -\cos(x) + 2\sin(x)$$
Show that the graph of $f(x) = x^3 + x^2 + x + 1$ has no horizontal tangent lines.
The first derivative of the function is
$$f'(x) = 3x^2 + 2x + 1$$For the graph to have a horizontal tangent line, whose slope would be 0, there must be at least one value of x for which $f'(x) = 0$. But the equation
$$3x^2 + 2x + 1 = 0$$doesn't have any solutions so there can't be any points with this kind of tangent line.
Find values of a, b and c such that $f(x) = ax^2 + bx + c$ is tangent to $y = 4x - 5$ at $x = 0$ and $y=-4x - 13$ at $x=-2$.
First, note that, since both the function and the tangent line touch at $x = 0$ and the corresponding y-coordinate of that tangent line is $y=4(0) - 5 = -5$ the quadratic function must also equal -5 at $x=0$. This gives us
$$-5 = f(0) = a(0)^2 + b(0) + c$$ $$c = -5$$This makes the equation, so far, $f(x) = ax^2 + bx - 5$.
Now, look at the two tangent lines. They tell us that the slope of the tangent line at $x = 0$ is 4 and the slope of the tangent line at $x = -2$ is -4. Since
$$f'(x) = 2ax + b$$Subsituting the slopes and x-values taken from the tangent lines gives us
$$f'(0) = 4$$ $$2a(0) + b = 4$$ $$b = 4$$and
$$f'(-2) = -4$$ $$2ax + b = -4$$ $$2a(-2) + 4 = -4$$ $$-4a = -8$$ $$a = 2$$So the equation of the original line must be $f(x) = 2x^2 + 4x - 5$.
Find values of a, b and c such that $f(x) = ax^3 + bx + c$ is tangent to $y = 13$ at $x = -1$ and $y=13x - 7$ at $x=1$.
$$f(x) = 3x^3 + 4x - 1$$
Find the equation of the normal line to $f(x) = 2\cos(x)$ at $x=\pi/4$.
$$f'(x) = -2\sin(x)$$ $$f'(\pi/4) = -\sqrt{2}$$
The slope of the normal line is the negative reciprocal of the value so $m_{\perp} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Since $y=f(\pi/4) = \sqrt{2}$, the equation of the normal line is
$$y - \sqrt{2} = \frac{\sqrt{2}}{2}(x - \pi/4)$$ $$y = \frac{\sqrt{2}}{2}x - \frac{\pi\sqrt{2}}{8} + \sqrt{2}$$Find the equation of the normal line to $f(x) = \cos(x)$ at $x=0$.
$$f'(x) = -\sin(x)$$ $$f'(0) = 0$$
Because the tangent line is horizontal, the line perpendicular to it must be vertical. Since it goes through the point $(0, 1)$, its equation must be $x=0$.
Find the equation of the normal line to $f(x) = x^3+3x^2-1$ at $x=1$.
$$y = -\frac{1}{9}x + \frac{10}{9}$$
