Week 6 Activity

Section 6.4 Week 6 Activity

Subsection 6.4.1 Derivatives By Definition

Activity 6.4.1.

Calculate this derivative by definition.

\begin{equation*} \frac{d}{dx} (x^2-4) \end{equation*}

Solution.

I use the limit definition of the derivative.

\begin{equation*} \frac{d}{dx} (x^2-4) = \lim_{h \rightarrow 0} \frac{(x+h)^2 - 4 - (x^2-4)}{h} \end{equation*}

Then I simplify this expression, trying to find a way to factor an \(h\) from the numerator so that I can cancel it with the \(h\) from the denominator.

\begin{align*} \lim_{h \rightarrow 0} \frac{(x+h)^2 - 4 - (x^2-4)}{h} \amp = \lim_{h \rightarrow 0} \frac{x^2 + 2xh + h^2 - 4 - x^2 + 4}{h} \\ \amp = \lim_{h \rightarrow 0} \frac{2xh + h^2}{h} \\ \amp = \lim_{h \rightarrow 0} \frac{h(2x + h)}{h} \\ \amp = \lim_{h \rightarrow 0} 2x + h = 2x+0 = 2x \end{align*}

Therefore, the derivative is \(2x\text{.}\)

Activity 6.4.2.

Calculate this derivative by definition.

\begin{equation*} \frac{d}{dx} \sqrt{5x} \end{equation*}

Solution.

I use the limit definition of the derivative.

\begin{equation*} \frac{d}{dx} \sqrt{5x} = \lim_{h \rightarrow 0} \frac{\sqrt{5(x+h)} - \sqrt{5x}}{h} \end{equation*}

Then I simplify this expression, trying to find a way to factor an \(h\) from the numerator so that I can cancel it with the \(h\) from the denominator. I use a trick here called multiplying by the conjugate. This trick will remove the square roots from the numerator (it also makes the denominator more complicated, but that’s not a problem). This trick is based on the idea that if I multiply the top and bottom of a fraction by the same thing, then I haven’t changed the value of the fraction at all.

\begin{align*} \lim_{h \rightarrow 0} \frac{\sqrt{5(x+h)} - \sqrt{5x}}{h} \amp = \lim_{h \rightarrow 0} \frac{(\sqrt{5(x+h)} - \sqrt{5x})(\sqrt{5(x+h)} + \sqrt{5x})}{h(\sqrt{5(x+h)} + \sqrt{5x})}\\ \amp = \lim_{h \rightarrow 0} \frac{5(x+h) - 5x}{h(\sqrt{5(x+h)} + \sqrt{5x})} \\ \amp = \lim_{h \rightarrow 0} \frac{5x + 5h - 5x}{h(\sqrt{5(x+h)} + \sqrt{5x})} \\ \amp = \lim_{h \rightarrow 0} \frac{5h}{h(\sqrt{5(x+h)} + \sqrt{5x})} \\ \amp = \lim_{h \rightarrow 0} \frac{5}{(\sqrt{5(x+h)} + \sqrt{5x})} \end{align*}

After multiplying by the conjugate, simplifying the numerator, and cancelling the \(h\) from numerator and denominator, I reach a point where I can simply evaluate the limit by replacing \(h\) with 0.

\begin{equation*} \lim_{h \rightarrow 0} \frac{5}{(\sqrt{5(x+h)} + \sqrt{5x})} = \frac{5}{\sqrt{5x}+ \sqrt{5x}} = \frac{5}{2\sqrt{5x}} \end{equation*}

This is a reasonable expression for the derivative of the original function.

Subsection 6.4.2 Derivatives

Activity 6.4.3.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} 8x^3 - 4x^2 + 3x + 1 \end{equation*}

Solution.

I can use linearity to split this up into four derivatives.

\begin{equation*} \frac{d}{dx} 8x^3 - 4x^2 + 3x + 1 = \frac{d}{dx} 8x^3 - \frac{d}{dx} 4x^2 + \frac{d}{dx} 3x + \frac{d}{dx} 1 \end{equation*}

I can use linearity again to take out the constant.

\begin{equation*} = 8 \frac{d}{dx} x^3 - 4 \frac{d}{dx} x^2 + 3 \frac{d}{dx} x + \frac{d}{dx} 1 \end{equation*}

Then I apply the product rule to the first two derivatives, and known deriavtive for the second pair.

\begin{equation*} = 8 (3x^2) - 4 (2x) + 3 (1) + 0 = 24x^2 - 8x + 3 \end{equation*}

Activity 6.4.4.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} 9x^4 - 2x^3 - x - 17 \end{equation*}

Solution.

I can use linearity to split this up and pull out the constant. (In the previous question, I did this over two linearity steps. As I move on, I tend to show fewer steps and apply the rules more quickly and less explicitly.)

\begin{equation*} \frac{d}{dx} 9x^4 - 2x^3 - x - 17 = 9 \frac{d}{dx} x^4 - 2 \frac{d}{dx} x^3 - \frac{d}{dx} x - \frac{d}{dx} 17 \end{equation*}

Then I use the power rule and known derivatives.

\begin{equation*} = 9 (4x^3) - 2 (3x^2) - 1 - 0 = 36x^3 - 6x^2 - 1 \end{equation*}

Activity 6.4.5.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} (x^2) \sin x \end{equation*}

Solution.

I use the product rule. I will label the pieces as \(f(x) = x^2\) and \(g(x) = \sin x\text{.}\) Here is the form of the product rule.

\begin{equation*} \frac{d}{dx} f(x) g(x) = f(x) \frac{d}{dx} g(x) + g(x) \frac{d}{dx} f(x) \end{equation*}

I apply this form to the piece of this product.

\begin{equation*} \frac{d}{dx} (x^2) \sin x = x^2 \frac{d}{dx} (\sin x) + (\sin x) \frac{d}{dx} x^2 \end{equation*}

I apply the power rule for the polynomial derivative and the known derivative of the sine function from the tables.

\begin{equation*} = x^2 (\cos x) + (\sin x) (2x) \end{equation*}

This is a reasonable way to leave the answer. However, by pure convention, I like to write polynoials before trig function, so I’ll write the answer slightly differently.

\begin{equation*} \frac{d}{dx} (x^2) \sin x = x^2\cos x + 2x\sin x \end{equation*}

Activity 6.4.6.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} (x^2+3)(x^3 - 7x - 1) \end{equation*}

Solution.

I use the product rule. I will label the pieces as \(f(x) = (x^2 + 3) \) and \(g(x) = (x^3 - 7x - 1)\text{.}\) Here is the form of the product rule.

\begin{equation*} \frac{d}{dx} f(x) g(x) = f(x) \frac{d}{dx} g(x) + g(x) \frac{d}{dx} f(x) \end{equation*}

I apply this form to the piece of this product.

\begin{equation*} \frac{d}{dx} (x^2+3)(x^3 - 7x - 1) = (x^2 + 3) \frac{d}{dx} (x^3 - 7x - 1) + (x^3 - 7x - 1) \frac{d}{dx} (x^2+3) \end{equation*}

Then, in each of the two derivatives, I use linearity to split the derivative up over the pieces.

\begin{equation*} = (x^2+3) \left( \frac{d}{dx} x^3 - 7 \frac{d}{dx} x - \frac{d}{dx} 1 \right) + (x^3 - 7x - 1) \left( \frac{d}{dx} x^2 + \frac{d}{dx} 3 \right) \end{equation*}

Then I use the power rule and known derivatives to evaluate all the individual derivatives.

\begin{equation*} = (x^2+3) \left( 3x^2 - 7 - 0 \right) + (x^3 - 7x - 1) \left( 2x + 0 \right) \end{equation*}

I could leave it in this form -- this is a valid answer. However, expanding and simplifying into a single polynomial is certainly a cleaner way to leave the answer.

\begin{equation*} = 3x^4 + 9x^2 - 7x^2 - 21 + 2x^4 - 14x^2 - 2x = 5x^4 - 12x^2 - 2x - 21 \end{equation*}

Activity 6.4.7.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \frac{x^2+3}{x-4} \end{equation*}

Solution.

I use the quotient rule. I will label the pieces as \(f(x) = x^2+3\) and \(g(x) = x-4\text{.}\) Here is the form of the quotient rule.

\begin{equation*} \frac{d}{dx} f(x) g(x) = \frac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x)}{(g(x))^2} \end{equation*}

I apply this form to the piece of this product.

\begin{equation*} \frac{(x-4) \frac{d}{dx} (x^2+3) - (x^2+3) \frac{d}{dx} (x-4)}{(x-4)^2} \end{equation*}

I use linearity to split up the derivatives in the numerator. Then I use the power rule and known derivatives for these derivatives.

\begin{equation*} \frac{(x-4) \left( 2x + 0 \right) - (x^2+3) (1-0)}{(x-4)^2} \end{equation*}

To put the answer in a more pleasant form, I distribute the multiplications in the numerator to write it as one polynomial.

\begin{equation*} \frac{2x^2 - 8x - x^2 - 3}{(x-4))^2} = \frac{x^2 - 8x - 3}{(x-4)^2} \end{equation*}

Activity 6.4.8.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \sqrt{x} + x^2 \end{equation*}

Solution.

I use linearity to split up the derivative.

\begin{equation*} \frac{d}{dx} \sqrt{x} + x^2 = \frac{d}{dx} \sqrt{x} + \frac{d}{dx} x^2 \end{equation*}

The second derivative is a power rule, which is hopefully pretty familiar with by this point. The first is also a power rule, if I write the square root as a fractional exponent: \(\sqrt{x} = x^\frac{1}{2}\text{.}\)

\begin{equation*} = \frac{d}{dx} x^{\frac{1}{2}} + \frac{d}{dx} x^2 = \frac{1}{2} x^{-\frac{1}{2}} + 2x \end{equation*}

To finish, I’ll return the fractional exponent to a square root form, since that is the more conventional notation.

\begin{equation*} = \frac{1}{2\sqrt{x}} + 2x \end{equation*}

Activity 6.4.9.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} e^x \sqrt{x} \end{equation*}

Solution.

I use the product rule. I will label the pieces as \(f(x) = e^x\) and \(g(x) = \sqrt{x}\text{.}\) Here is the form of the product rule.

\begin{equation*} \frac{d}{dx} f(x) g(x) = f(x) \frac{d}{dx} g(x) + g(x) \frac{d}{dx} f(x) \end{equation*}

I apply this form to the piece of this product.

\begin{equation*} \frac{d}{dx} e^x \sqrt{x} = e^x \frac{d}{dx} \sqrt{x} + \sqrt{x} \frac{d}{dx} e^x \end{equation*}

The derivative of the exponential function is itself, but we have to write the square root as a fracitonal exponent to use the power rule to differentiate it.

\begin{equation*} = e^x \frac{d}{dx} x^\frac{1}{2} + \sqrt{x} \frac{d}{dx} e^x = e^x \frac{1}{2} x^{\frac{-1}{2}} + \sqrt{x} e^x \end{equation*}

I’ll change the fractional exponent back to a conventional square root expression. I’ll also factor out the exponential term which is common to both expressions.

\begin{equation*} = e^x \left( \frac{1}{2\sqrt{x}} + \sqrt{x} \right) \end{equation*}

Activity 6.4.10.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \frac{\sqrt{x}}{x^2+4} \end{equation*}

Solution.

I use the quotient rule. I will label the pieces as \(f(x) = \sqrt{x}\) and \(g(x) = x^2+4\text{.}\) Here is the form of the product rule.

\begin{equation*} \frac{d}{dx} f(x) g(x) = g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x) \end{equation*}

I apply this form to the piece of this product.

\begin{equation*} \frac{d}{dx} \frac{\sqrt{x}}{x^2+4} = \frac{ (x^2+4) \frac{d}{dx} \sqrt{x} - \sqrt{x} \frac{d}{dx} (x^2+4)}{(x^2+4)^2} \end{equation*}

I need to evaluate both the derivative in the numerator. The second can be evaluated with linearity and the power rule, like all polynomials. The first is evaluated by writing the square root as a fractional exponent.

\begin{align*} \amp = \frac{ (x^2+4) \frac{d}{dx} x^\frac{1}{x} - \sqrt{x} \left( \frac{d}{dx} x^2 + \frac{d}{dx} 4 \right)}{(x^2+4)^2} \\ \amp = \frac{ (x^2+4) \frac{1}{2} x^{\frac{-1}{2}} - \sqrt{x} (2x+0)}{(x^2+4)^2} \end{align*}

This is an answers, but as usual, I’ll make some simplifications to give a more pleasant and usable form of the answer.

\begin{equation*} = \frac{\frac{x^2+4}{2\sqrt{x}} - 2x\sqrt{x}}{(x^2+4)^2} = \frac{x^2 + 4 - 4x^2}{2(x^2+4)\sqrt{x}} = \frac{-3x^2+4}{2(x^2+4)\sqrt{x}} \end{equation*}

Activity 6.4.11.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \frac{e^x}{x^2+1} \end{equation*}

Solution.

I use the quotient rule. I will label the pieces as \(f(x) = e^x\) and \(g(x) = x^2+1\text{.}\) Here is the form of the quotient rule.

\begin{equation*} \frac{d}{dx} f(x) g(x) = \frac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x)}{(g(x))^2} \end{equation*}

I apply this form to the piece of this product.

\begin{equation*} \frac{d}{dx} \frac{e^x}{x^2+1} = \frac{ (x^2+1) \frac{d}{dx} e^x - e^x \frac{d}{dx} (x^2+1)}{(x^2+1)^2} \end{equation*}

The first derivative is the known derivative of the exponential function. The second deriative, like all polynomials, is solved by linearity and the power rule.

\begin{equation*} = \frac{d}{dx} \frac{e^x}{x^2+1} = \frac{ (x^2+1) e^x - e^x (2x+0)}{(x^2+1)^2} \end{equation*}

As usual, I’ll make some simplifications to the result of the derivative.

\begin{equation*} \frac{d}{dx} \frac{e^x}{x^2+1} = \frac{ e^x (x^2 - 2x +1)} {(x^2+1)^2} \end{equation*}

Activity 6.4.12.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \frac{\cos x}{x} + \frac{\sin x}{x} \end{equation*}

Solution.

These are both quotient rule derivatives. I’ll not write the general form and labelling here, but proceed with the specific calculation.

\begin{equation*} \frac{d}{dx} \frac{\cos x}{x} + \frac{\sin x}{x} = \frac{x \frac{d}{dx} \cos x - \cos x \frac{d}{dx} x}{x^2} + \frac{x \frac{d}{dx} \sin x - \sin x \frac{d}{dx} x}{x^2} \end{equation*}

Then I need to do all four of the specific derivatives, which are all known derivatives.

\begin{equation*} = \frac{-x \sin x - \cos x}{x^2} + \frac{x \cos x - \sin x}{x^2} \end{equation*}

Some change in the expression could be made — perhaps splitting up the numerators. No change seems to me to create any substantially simpler expression, so I’m content to leave the answer in this form.

Activity 6.4.13.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} (x^2+1)(\sin x)(e^x) \end{equation*}

Solution.

This is a triple product. I still need the product rule, but I have to be conscious of the grouping. There are several options, but I will choose \(f(x) = (x^2+1)\) and \(g(x) = (\sin x)(e^x)\text{.}\) Then I proceed with the product rule form.

\begin{equation*} \frac{d}{dx} (x^2+1)(\sin x)(e^x) = (x^2+1) \frac{d}{dx} \left( (\sin x)(e^x) \right) + (\sin x)(e^x) \frac{d}{dx} (x^2+1) \end{equation*}

The second derivative is a reasonable polynomial, solved by linearity. The first derivative, however, is still a product. I need to do another product rule with this product.

\begin{equation*} = (x^2+1) \left( e^x \frac{d}{dx} \sin x + \sin x \frac{d}{dx} e^x \right) + (\sin x)(e^x) (2x+0) \end{equation*}

Then I just need to use the two known derivatives of the sine function and the exponential function.

\begin{equation*} = (x^2+1) \left( e^x \cos x + \sin x e^x \right) + (\sin x)(e^x) (2x) \end{equation*}

I can simplify this slightly by factoring out the exponential and combining the polynomial terms.

\begin{align*} \amp = e^x \left( (x^2+1) \cos x + (x^2+1) \sin x + (2x) (\sin x) \right) \\ \amp = e^x \left( (x^2+1) \cos x + (x^2 + 2x + 1) \sin x \right) \end{align*}

Subsection 6.4.3 Conceptual Review Questions

What is a derivative? What does it do and what does it mean?
Why is a derivative calculated by a limit? Why do I need limits to define derivatives?
What is linearity? Why is the derivative linear?
What are the product and quotient rules? Why isn’t the derivative of a product or quotient simply the product or quotient of the derivatives?
Why are square root and other roots the same as fractional exponents?

Prev Top Next