These derivatives practice problems are organized progressively from foundational differentiation rules to advanced techniques including the chain rule, implicit differentiation, and higher-order derivatives. Whether you are building fluency with the power rule or tackling composite functions and logarithmic differentiation, each section below provides structured exercises with hints and fully worked solutions. This resource covers every core topic tested in Calculus I, from the limit definition of the derivative to applications such as tangent lines and rates of change.
Power Rule and Basic Differentiation
The power rule is the cornerstone of differentiation. Given \( f(x) = x^n \), we have \( f'(x) = nx^{n-1} \). Mastery here is essential before progressing to more advanced rules.
Exercise 1: Differentiating Polynomial Functions
Easy
Consider the polynomial function \( f(x) = 5x^4 – 3x^3 + 7x^2 – 2x + 9 \).
- Find \( f'(x) \) using the power rule applied term by term.
- Evaluate \( f'(1) \) and interpret its meaning geometrically.
Hint
View Solution
\[ f'(x) = 20x^3 – 9x^2 + 14x – 2 \]
Solution to question 2:
\[ f'(1) = 20(1)^3 – 9(1)^2 + 14(1) – 2 = 20 – 9 + 14 – 2 = 23 \]
The slope of the tangent line to \( f \) at \( x = 1 \) is \( 23 \).
Exercise 2: Negative and Fractional Exponents
Easy
Rewrite the following function using exponent notation, then differentiate:
\[ g(x) = \frac{6}{\sqrt{x^3}} + \frac{4}{x^2} – 3\sqrt{x} \]
- Express each term as a power of \( x \) and find \( g'(x) \).
- Simplify \( g'(x) \) completely.
Hint
View Solution
\[ g(x) = 6x^{-3/2} + 4x^{-2} – 3x^{1/2} \]
\[ g'(x) = 6 \cdot \left(-\frac{3}{2}\right)x^{-5/2} + 4 \cdot (-2)x^{-3} – 3 \cdot \frac{1}{2}x^{-1/2} \]
Solution to question 2:
\[ g'(x) = -9x^{-5/2} – 8x^{-3} – \frac{3}{2}x^{-1/2} = -\frac{9}{x^{5/2}} – \frac{8}{x^3} – \frac{3}{2\sqrt{x}} \]
Exercise 3: Tangent Line to a Curve
Medium
Let \( h(x) = 2x^3 – 5x + 1 \).
- Find the equation of the tangent line to \( h \) at \( x = 2 \).
- Determine all values of \( x \) where the tangent line is horizontal.
Hint
View Solution
\[ h'(x) = 6x^2 – 5 \]
\[ h'(2) = 6(4) – 5 = 19, \quad h(2) = 16 – 10 + 1 = 7 \]
\[ \text{Tangent line: } y – 7 = 19(x – 2) \implies y = 19x – 31 \]
Solution to question 2:
\[ 6x^2 – 5 = 0 \implies x^2 = \frac{5}{6} \implies x = \pm\sqrt{\frac{5}{6}} \approx \pm 0.913 \]
Product Rule and Quotient Rule
When two differentiable functions are multiplied or divided, the product and quotient rules provide the systematic approach. These rules are frequently tested and appear throughout physics, engineering, and economics.
Exercise 4: Applying the Product Rule
Easy
Differentiate the following functions using the product rule \( (fg)’ = f’g + fg’ \).
- \( p(x) = (3x^2 – 1)(x^3 + 4x) \)
- \( q(x) = x^5 \sin x \)
Hint
View Solution
\[ p'(x) = (6x)(x^3 + 4x) + (3x^2 – 1)(3x^2 + 4) \]
\[ = 6x^4 + 24x^2 + 9x^4 + 12x^2 – 3x^2 – 4 = 15x^4 + 33x^2 – 4 \]
Solution to question 2:
\[ q'(x) = 5x^4 \sin x + x^5 \cos x \]
Exercise 5: Applying the Quotient Rule
Easy
Differentiate each function using the quotient rule \( \left(\dfrac{f}{g}\right)’ = \dfrac{f’g – fg’}{g^2} \).
- \( r(x) = \dfrac{x^2 + 3}{2x – 5} \)
- \( s(x) = \dfrac{\cos x}{x^3} \)
Hint
View Solution
\[ r'(x) = \frac{(2x)(2x-5) – (x^2+3)(2)}{(2x-5)^2} = \frac{4x^2 – 10x – 2x^2 – 6}{(2x-5)^2} = \frac{2x^2 – 10x – 6}{(2x-5)^2} \]
Solution to question 2:
\[ s'(x) = \frac{-\sin x \cdot x^3 – \cos x \cdot 3x^2}{x^6} = \frac{-x^2(x\sin x + 3\cos x)}{x^6} = \frac{-(x\sin x + 3\cos x)}{x^4} \]
Exercise 6: Combined Product and Quotient Rules
Medium
A particle’s position is given by \( s(t) = \dfrac{t^2 + 1}{e^t} \), where \( t \geq 0 \).
- Find the velocity \( v(t) = s'(t) \).
- Determine the value(s) of \( t \) where the particle is momentarily at rest.
Hint
View Solution
\[ v(t) = \frac{2t \cdot e^t – (t^2+1) \cdot e^t}{e^{2t}} = \frac{e^t(2t – t^2 – 1)}{e^{2t}} = \frac{-t^2 + 2t – 1}{e^t} = \frac{-(t-1)^2}{e^t} \]
Solution to question 2:
\[ v(t) = 0 \implies -(t-1)^2 = 0 \implies t = 1 \]
The particle is momentarily at rest at \( t = 1 \).
Chain Rule and Composite Functions
The chain rule handles composite functions of the form \( f(g(x)) \). It is among the most widely applied differentiation rules: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). Identifying the outer and inner functions is the key first step.
Exercise 7: Chain Rule — Polynomial Inside a Power
Easy
Differentiate the following functions using the chain rule.
- \( f(x) = (4x^2 – 3x + 1)^6 \)
- \( g(x) = \sqrt{7x^3 – 2x} \)
Hint
View Solution
\[ f'(x) = 6(4x^2 – 3x + 1)^5 \cdot (8x – 3) \]
Solution to question 2:
\[ g(x) = (7x^3 – 2x)^{1/2} \]
\[ g'(x) = \frac{1}{2}(7x^3 – 2x)^{-1/2} \cdot (21x^2 – 2) = \frac{21x^2 – 2}{2\sqrt{7x^3 – 2x}} \]
Exercise 8: Chain Rule with Trigonometric Functions
Medium
Find the derivative of each function.
- \( h(x) = \sin(5x^2 – 3x) \)
- \( k(x) = \cos^4(x) \) — note this means \( [\cos(x)]^4 \)
Hint
View Solution
\[ h'(x) = \cos(5x^2 – 3x) \cdot (10x – 3) \]
Solution to question 2:
\[ k'(x) = 4\cos^3(x) \cdot (-\sin x) = -4\sin x \cos^3 x \]
Exercise 9: Nested Chain Rule
Hard
Differentiate the following functions, which require applying the chain rule more than once.
- \( m(x) = e^{\sin(x^3)} \)
- \( n(x) = \ln\!\left(\sqrt{1 + \cos^2 x}\right) \)
Hint
View Solution
\[ m'(x) = e^{\sin(x^3)} \cdot \cos(x^3) \cdot 3x^2 \]
Solution to question 2:
\[ n(x) = \frac{1}{2}\ln(1 + \cos^2 x) \]
\[ n'(x) = \frac{1}{2} \cdot \frac{1}{1 + \cos^2 x} \cdot 2\cos x \cdot (-\sin x) = \frac{-\sin x \cos x}{1 + \cos^2 x} \]
Derivatives of Exponential and Logarithmic Functions
Exponential and logarithmic derivatives are indispensable in modeling growth, decay, and numerous applied problems. Key facts: \( \frac{d}{dx}[e^x] = e^x \), \( \frac{d}{dx}[\ln x] = \frac{1}{x} \), and for general bases: \( \frac{d}{dx}[a^x] = a^x \ln a \).
Exercise 10: Exponential Derivatives
Easy
Find the derivative of each function.
- \( f(x) = 3e^x – 5e^{2x} \)
- \( g(x) = 4^x \cdot x^2 \)
Hint
View Solution
\[ f'(x) = 3e^x – 10e^{2x} \]
Solution to question 2:
\[ g'(x) = 4^x \ln 4 \cdot x^2 + 4^x \cdot 2x = 4^x(x^2 \ln 4 + 2x) \]
Exercise 11: Logarithmic Derivatives
Medium
Differentiate the following functions involving natural and general logarithms.
- \( p(x) = \ln(x^3 + 2x – 1) \)
- \( q(x) = \log_5(x^2 + 1) \)
Hint
View Solution
\[ p'(x) = \frac{3x^2 + 2}{x^3 + 2x – 1} \]
Solution to question 2:
\[ q(x) = \frac{\ln(x^2+1)}{\ln 5} \implies q'(x) = \frac{2x}{(x^2+1)\ln 5} \]
Exercise 12: Logarithmic Differentiation
Hard
Use logarithmic differentiation to find \( y’ \) for functions involving variable exponents or complex products.
- \( y = x^{\sin x} \)
- \( y = \dfrac{(x^2+1)^3 \sqrt{x+4}}{(2x-1)^5} \)
Hint
View Solution
\[ \ln y = \sin x \cdot \ln x \]
\[ \frac{y’}{y} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]
\[ y’ = x^{\sin x}\!\left(\cos x \ln x + \frac{\sin x}{x}\right) \]
Solution to question 2:
\[ \ln y = 3\ln(x^2+1) + \frac{1}{2}\ln(x+4) – 5\ln(2x-1) \]
\[ \frac{y’}{y} = \frac{6x}{x^2+1} + \frac{1}{2(x+4)} – \frac{10}{2x-1} \]
\[ y’ = \frac{(x^2+1)^3\sqrt{x+4}}{(2x-1)^5} \left(\frac{6x}{x^2+1} + \frac{1}{2(x+4)} – \frac{10}{2x-1}\right) \]
Implicit Differentiation
Not every relationship between \( x \) and \( y \) can be solved explicitly for \( y \). Implicit differentiation allows us to differentiate both sides of an equation with respect to \( x \), treating \( y \) as a function of \( x \) and applying the chain rule wherever \( y \) appears.
Exercise 13: Finding dy/dx Implicitly
Medium
Use implicit differentiation to find \( \dfrac{dy}{dx} \) for each equation.
- \( x^3 + y^3 = 6xy \)
- \( \sin(xy) = x + y \)
Hint
View Solution
\[ 3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx} \]
\[ (3y^2 – 6x)\frac{dy}{dx} = 6y – 3x^2 \]
\[ \frac{dy}{dx} = \frac{6y – 3x^2}{3y^2 – 6x} = \frac{2y – x^2}{y^2 – 2x} \]
Solution to question 2:
\[ \cos(xy)\!\left(y + x\frac{dy}{dx}\right) = 1 + \frac{dy}{dx} \]
\[ y\cos(xy) + x\cos(xy)\frac{dy}{dx} = 1 + \frac{dy}{dx} \]
\[ \frac{dy}{dx}(x\cos(xy) – 1) = 1 – y\cos(xy) \]
\[ \frac{dy}{dx} = \frac{1 – y\cos(xy)}{x\cos(xy) – 1} \]
Exercise 14: Tangent Line via Implicit Differentiation
Hard
Consider the lemniscate \( (x^2 + y^2)^2 = 4(x^2 – y^2) \).
- Find \( \dfrac{dy}{dx} \) implicitly.
- Find the equation of the tangent line at the point \( (\sqrt{2},\, 0) \).
Hint
View Solution
\[ 2(x^2+y^2)\left(2x + 2y\frac{dy}{dx}\right) = 4\left(2x – 2y\frac{dy}{dx}\right) \]
\[ 4(x^2+y^2)\!\left(x + y\frac{dy}{dx}\right) = 8x – 8y\frac{dy}{dx} \]
\[ 4x(x^2+y^2) + 4y(x^2+y^2)\frac{dy}{dx} = 8x – 8y\frac{dy}{dx} \]
\[ \frac{dy}{dx}\!\left[4y(x^2+y^2) + 8y\right] = 8x – 4x(x^2+y^2) \]
\[ \frac{dy}{dx} = \frac{8x – 4x(x^2+y^2)}{4y(x^2+y^2) + 8y} = \frac{x\left[2 – (x^2+y^2)\right]}{y\left[(x^2+y^2) + 2\right]} \]
Solution to question 2:
At \( (\sqrt{2}, 0) \): \( x^2+y^2 = 2 \). The expression has \( y = 0 \) in the denominator, so we return to the step before dividing:
\[ \frac{dy}{dx}\!\left[4y(x^2+y^2) + 8y\right] = 8x – 4x(x^2+y^2) \]
At \( (\sqrt{2}, 0) \): left side \( = 0 \) and right side \( = 8\sqrt{2} – 4\sqrt{2}(2) = 8\sqrt{2} – 8\sqrt{2} = 0 \). Since both sides are zero for all \( \frac{dy}{dx} \), the tangent is vertical: the tangent line is \( x = \sqrt{2} \).
Higher-Order Derivatives and Their Applications
The second derivative \( f”(x) \) measures the rate of change of the first derivative — describing concavity and acceleration. Higher-order derivatives appear in physics, Taylor series, and curve analysis.
Exercise 15: Computing Second and Third Derivatives
Easy
Find the requested higher-order derivative for each function.
- Find \( f”(x) \) for \( f(x) = 4x^5 – 3x^4 + x^2 – 7 \).
- Find \( g”'(x) \) for \( g(x) = \sin(2x) \).
Hint
View Solution
\[ f'(x) = 20x^4 – 12x^3 + 2x \]
\[ f”(x) = 80x^3 – 36x^2 + 2 \]
Solution to question 2:
\[ g'(x) = 2\cos(2x),\quad g”(x) = -4\sin(2x),\quad g”'(x) = -8\cos(2x) \]
Exercise 16: Concavity and Inflection Points
Medium
Let \( f(x) = x^4 – 8x^2 + 3 \).
- Find \( f”(x) \) and determine the intervals where \( f \) is concave up and concave down.
- Identify all inflection points of \( f \).
Hint
View Solution
\[ f'(x) = 4x^3 – 16x,\quad f”(x) = 12x^2 – 16 \]
\[ f”(x) = 0 \implies x^2 = \frac{4}{3} \implies x = \pm \frac{2}{\sqrt{3}} \]
Concave up: \( \left(-\infty,\,-\dfrac{2}{\sqrt{3}}\right) \cup \left(\dfrac{2}{\sqrt{3}},\,+\infty\right) \). Concave down: \( \left(-\dfrac{2}{\sqrt{3}},\,\dfrac{2}{\sqrt{3}}\right) \).
Solution to question 2:
At \( x = \pm\dfrac{2}{\sqrt{3}} \), \( f” \) changes sign, so there are inflection points at:
\[ \left(\frac{2}{\sqrt{3}},\; f\!\left(\frac{2}{\sqrt{3}}\right)\right) \quad \text{and} \quad \left(-\frac{2}{\sqrt{3}},\; f\!\left(-\frac{2}{\sqrt{3}}\right)\right) \]
\[ f\!\left(\tfrac{2}{\sqrt{3}}\right) = \frac{16}{9} – \frac{32}{3} + 3 = \frac{16 – 96 + 27}{9} = -\frac{53}{9} \]
Inflection points: \( \left(\pm\dfrac{2}{\sqrt{3}},\,-\dfrac{53}{9}\right) \).
Exercise 17: Second Derivative of an Implicitly Defined Function
Hard
Given the ellipse \( x^2 + 4y^2 = 16 \):
- Find \( \dfrac{dy}{dx} \) using implicit differentiation.
- Find \( \dfrac{d^2y}{dx^2} \) by differentiating your result from part 1 implicitly. Express the answer in terms of \( x \) and \( y \) only, then substitute using the original equation to simplify.
Hint
View Solution
\[ 2x + 8y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{4y} \]
Solution to question 2:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(-\frac{x}{4y}\right) = -\frac{4y \cdot 1 – x \cdot 4\frac{dy}{dx}}{16y^2} \]
Substitute \( \frac{dy}{dx} = -\frac{x}{4y} \):
\[ = -\frac{4y – x \cdot \left(-\frac{x}{y}\right)}{16y^2} = -\frac{4y + \frac{x^2}{y}}{16y^2} = -\frac{4y^2 + x^2}{16y^3} \]
Since \( x^2 + 4y^2 = 16 \):
\[ \frac{d^2y}{dx^2} = -\frac{16}{16y^3} = -\frac{1}{y^3} \]
Derivatives of Trigonometric Functions
The six trigonometric functions have well-established derivatives. Fluency with these — particularly recognizing when to combine trigonometric rules with the chain rule or product rule — is essential for success in Calculus I and beyond.
| Function | Derivative |
|---|---|
| \( \sin x \) | \( \cos x \) |
| \( \cos x \) | \( -\sin x \) |
| \( \tan x \) | \( \sec^2 x \) |
| \( \csc x \) | \( -\csc x \cot x \) |
| \( \sec x \) | \( \sec x \tan x \) |
| \( \cot x \) | \( -\csc^2 x \) |
Exercise 18: Basic Trigonometric Derivatives
Easy
Differentiate each function.
- \( f(x) = 3\tan x – 2\csc x \)
- \( g(x) = x^2 \cos x \)
Hint
View Solution
\[ f'(x) = 3\sec^2 x + 2\csc x \cot x \]
Solution to question 2:
\[ g'(x) = 2x\cos x + x^2(-\sin x) = 2x\cos x – x^2\sin x \]
Exercise 19: Advanced Trigonometric Problem
Hard
Let \( f(x) = \dfrac{\sin x}{1 + \cos x} \).
- Find \( f'(x) \) and simplify as much as possible.
- Use a trigonometric identity to show that \( f'(x) = \dfrac{1}{1 + \cos x} \).
Hint
View Solution
\[ f'(x) = \frac{\cos x(1+\cos x) – \sin x(-\sin x)}{(1+\cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2} \]
Solution to question 2:
Using \( \sin^2 x + \cos^2 x = 1 \):
\[ f'(x) = \frac{\cos x + 1}{(1+\cos x)^2} = \frac{1 + \cos x}{(1+\cos x)^2} = \frac{1}{1+\cos x} \]