Test Your Knowledge: Derivatives

This page will serve as a practice test for derivatives. We will go over some common derivative problems and see fully worked solutions for how they are solved.

1. Determine the derivative of f(x) = x^2 + 2x +5 from first principles

When we say “first principles”, we mean to use the limit definition of a derivative. From the rules we learned in the previous sections, we know this derivative should be f(x) = 2x + 2, so let’s us the limit definition to prove it.

lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) + 5 - (x^2+2x+5)}{h}
=lim_{h \to 0} \frac{x^2+2xh+h^2 + 2x+2h + 5 - x^2-2x-5}{h}
=lim_{h \to 0} \frac{2xh+h^2+2h}{h}
=lim_{h \to 0} \frac{2xh+h^2+2h}{h}
=lim_{h \to 0}2x+h+2
=2x+2

2. Determine the derivative of f(x) = (sin(x+3))(cos(x^2+1))

This problem combines chain rule and product rule. In this instances, it is helpful to calculate the derivatives of our two functions, and then plug them into product rule.

Let g(x) = sin(x+3) and let h(x)  = cos(x^2+1)

g'(x) = cos(x+3)
h'(x) = -sin(x^2+1)(2x)

Now that we know the individual derivatives, product rule becomes as easy as plugging in the values.

f'(x) = g'(x)h(x) + g(x)h'(x)
f'(x) = cos(x+3)cos(x^2+1) + sin(x+3)(-sin(x^2+1)(2x))

3. Determine the derivative of f(x) = \frac{e^x}{(x+2)^{20}}

Similar to problem 2, we have a chain rule derivative in a quotient rule problem. We will again identify our functions, derivatives, and plug in to solve the problem.

Let g(x) = e^x and let h(x) = (x+2)^{20}

g'(x) = e^x
h'(x) = 20(x+2)^{19}

f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{[g(x)]^2}
f'(x) = \frac{e^x*(x+2)^{20} - (e^x)(20(x+2)^{19}}{[(x+2)^{20}]^2}

4. Determine \frac{dy}{dx}(cos^{-1}(x))

y = cos^-1(x) \Longrightarrow cos(y) = x

When we rearrange in this way, we can use implicit differentiation.

\frac{dy}{dx}(cos(y)) = \frac{dy}{dx}(x)
\Longrightarrow -sin(y) * \frac{dy}{dx} = 1
\Longrightarrow \frac{dy}{dx} = \frac{1}{-sin(y)}

Now we just need to find sin(y) and we are done

sin^2(y) + cos^2(y) = 1
\Longrightarrow sin^2(y) = 1 - cos^2(y)
\Longrightarrow sin(y) = \sqrt{1-cos^2(y)}

Since we know cos(y) = x, we get: \frac{dy}{dx} = \frac{1}{-\sqrt{1-x^2}}

5. Determine \frac{dy}{dx} for y^2 + 2x^2 = y^3 + 1

This is a typical implicit differentiation question. We can’t isolate for y, so we need to use implicit differentiation

\Longrightarrow 2y * \frac{dy}{dx} + 4x = 3y^2 * \frac{dy}{dx} + 1
\Longrightarrow 2y * \frac{dy}{dx} - 3y^2 * \frac{dy}{dx} = 1-4x
\Longrightarrow (2y-3y^2) \frac{dy}{dx} = 1-4x
\Longrightarrow \frac{dy}{dx} \frac{1-4x}{2y-3y^2}

From these problems, there are a few key takeaways to remember when solving derivative problems.

  1. For problems involving first principles, don’t be afraid to verify your answer using the derivative rules you already know.
  2. When problems combine rules like chain and product rule, it can be valuable to calculate the derivatives of the individual functions before trying to put them through product and quotient rule.
  3. When you encounter functions that are inverses, try converting them into a form that can benefit from implicit differentiation.

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