Mathematics

Differentiation and Derivatives

55 Questions

Differentiation involves finding the rate of change of functions using established mathematical rules. Topics include the chain rule, quotient rule, and calculating partial derivatives for multivariable equations. Mastery of these mathematical calculations is heavily tested in undergraduate entrance and competitive exams.

Chain rule derivativesQuotient rule applicationPartial derivatives calculationDirectional derivative vectorsLogarithmic function derivatives

Differentiation and Derivatives Questions

Multiple choice

What is the derivative of $f(x) = e^(2x - 1)$ using the chain rule?

  1. 2e^(2x - 1)

  2. e^(2x - 1)

  3. 2e^(2x)

  4. e^(2x - 2)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the derivative of $e^(2x - 1)$ using the chain rule, we first identify the outer function $g(u) = e^u$ and the inner function $h(x) = 2x - 1$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = e^u$ and $h'(x) = 2$. Plugging these values in, we get $f'(x) = e^(2x - 1) * 2 = 2e^(2x - 1)$.

Multiple choice

Find the derivative of $f(x) = ln(x^2 + 1)$ using the chain rule.

  1. \frac{2x}{x^2 + 1}

  2. \frac{1}{x^2 + 1}

  3. \frac{1}{2x}

  4. \frac{2x}{2x^2 + 1}
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the derivative of $ln(x^2 + 1)$ using the chain rule, we first identify the outer function $g(u) = ln(u)$ and the inner function $h(x) = x^2 + 1$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = \frac{1}{u}$ and $h'(x) = 2x$. Plugging these values in, we get $f'(x) = \frac{1}{x^2 + 1} * 2x = \frac{2x}{x^2 + 1}$.

Multiple choice

What is the derivative of $f(x) = \frac{sin(x)}{cos(x)}$ using the quotient rule?

  1. \frac{cos(x) - sin(x)}{cos^2(x)}

  2. \frac{cos(x) + sin(x)}{cos^2(x)}

  3. \frac{sin(x)}{cos^2(x)}

  4. \frac{cos(x)}{sin^2(x)}
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the derivative of $\frac{sin(x)}{cos(x)}$ using the quotient rule, we first identify the two functions $g(x) = sin(x)$ and $h(x) = cos(x)$. Then, we apply the quotient rule formula: $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}$. In this case, $g'(x) = cos(x)$ and $h'(x) = -sin(x)$. Plugging these values in, we get $f'(x) = \frac{cos(x)cos(x) - sin(x)(-sin(x))}{cos^2(x)} = \frac{cos^2(x) + sin^2(x)}{cos^2(x)} = \frac{1}{cos^2(x)} = \frac{cos(x) - sin(x)}{cos^2(x)}$.

Multiple choice

Find the derivative of $f(x) = tan(x)$ using the chain rule.

  1. sec^2(x)

  2. tan(x)

  3. sec(x)

  4. 1 + tan^2(x)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the derivative of $tan(x)$ using the chain rule, we first identify the outer function $g(u) = tan(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = sec^2(u)$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = sec^2(x) * 1 = sec^2(x)$.

Multiple choice

What is the derivative of $f(x) = cot(x)$ using the chain rule?

  1. -csc^2(x)

  2. csc^2(x)

  3. cot(x)

  4. 1 - cot^2(x)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the derivative of $cot(x)$ using the chain rule, we first identify the outer function $g(u) = cot(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = -csc^2(u)$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = -csc^2(x) * 1 = -csc^2(x)$.

Multiple choice

Find the derivative of $f(x) = arcsin(x)$ using the chain rule.

  1. \frac{1}{\sqrt{1 - x^2}}

  2. \frac{1}{x^2}

  3. \frac{1}{1 - x^2}

  4. arcsin(x)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the derivative of $arcsin(x)$ using the chain rule, we first identify the outer function $g(u) = arcsin(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = \frac{1}{\sqrt{1 - u^2}}$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = \frac{1}{\sqrt{1 - x^2}} * 1 = \frac{1}{\sqrt{1 - x^2}}$.

Multiple choice

What is the derivative of the function $f(x) = x^3 - 2x^2 + 3x - 4$?

  1. $3x^2 - 4x + 3$
  2. $3x^2 - 4x - 4$
  3. $3x^2 - 4x + 4$
  4. $3x^2 - 4x - 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative of a function $f(x)$ is given by the formula $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$. Applying this formula to the given function, we get $f'(x) = \lim_{h \to 0} \frac{(x + h)^3 - 2(x + h)^2 + 3(x + h) - 4 - (x^3 - 2x^2 + 3x - 4)}{h} = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 2(x^2 + 2xh + h^2) + 3x + 3h - 4 - x^3 + 2x^2 - 3x + 4}{h} = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 - 4xh - 4h^2 + 3h}{h} = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 - 4x - 4h + 3)}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2 - 4x - 4h + 3) = 3x^2 - 4x + 3$.

Multiple choice

Find the derivative of the function (f(x) = x^3 - 2x^2 + 3x - 4).

  1. \(3x^2 - 4x + 3\)
  2. \(3x^2 - 2x + 3\)
  3. \(x^3 - 4x + 3\)
  4. \(x^3 - 2x^2 + 3\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the power rule of differentiation, we get (f'(x) = \frac{d}{dx}(x^3 - 2x^2 + 3x - 4) = 3x^2 - 2(2x) + 3(1) - 0 = 3x^2 - 4x + 3).

Multiple choice

What is the derivative of the function (f(x) = \sin(x^2 + 1))?

  1. \(2x\cos(x^2 + 1)\)
  2. \(x\cos(x^2 + 1)\)
  3. \(2x\sin(x^2 + 1)\)
  4. \(x\sin(x^2 + 1)\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the chain rule, we get (f'(x) = \frac{d}{dx}[\sin(x^2 + 1)]\frac{d}{dx}[x^2 + 1] = \cos(x^2 + 1)(2x) = 2x\cos(x^2 + 1)).

Multiple choice

What is the derivative of the function f(x) = x^3 - 2x^2 + 3x - 4?

  1. f'(x) = 3x^2 - 4x + 3

  2. f'(x) = 3x^2 - 2x + 3

  3. f'(x) = 3x^2 - 2x + 1

  4. f'(x) = 3x^2 - 4x + 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative of a function f(x) is given by the formula f'(x) = lim_(h->0) [f(x+h) - f(x)] / h. Substituting the values of f(x) and f(x+h), we get f'(x) = lim_(h->0) [(x+h)^3 - 2(x+h)^2 + 3(x+h) - 4 - (x^3 - 2x^2 + 3x - 4)] / h. Simplifying this equation, we get f'(x) = lim_(h->0) [3x^2 + 6xh + 3h^2 - 4x - 4h + 3 - x^3 + 2x^2 - 3x + 4] / h. Dividing both the numerator and denominator by h, we get f'(x) = lim_(h->0) [3x^2 + 6x + 3h - 4 - x^3 + 2x^2 - 3x + 4] / 1. Simplifying this equation further, we get f'(x) = 3x^2 - 4x + 3.