Derivatives: rules and functions

Normal lines

The normal line at a point on a curve is the line perpendicular to the tangent there, and when f′(a)≠0f'(a)\ne 0 its slope is −1f′(a)-\frac{1}{f'(a)}.

y−f(a)=−1f′(a)(x−a)(f′(a)≠0)\begin{gathered} y-f(a) \\[4pt] =-\frac{1}{f'(a)}(x-a)\quad(f'(a)\ne 0) \end{gathered}

The slopes of two perpendicular lines multiply to −1-1. The tangent has slope f′(a)f'(a), so the normal slope mm satisfies f′(a)⋅m=−1f'(a)\cdot m=-1, giving m=−1f′(a)m=-\frac{1}{f'(a)}.

y−f(a)=−1f′(a)(x−a)y-f(a)=-\frac{1}{f'(a)}(x-a)

For example, for y=x2y=x^2 at x=2x=2, f′(2)=4f'(2)=4, so the normal slope is −14-\frac{1}{4}. The smaller the absolute value of the tangent slope, the larger the absolute value of the normal slope.

Examples

  1. 01What is the slope of the normal line at this point?f′(a)=3\displaystyle f'(a)=3AnswerHide
    Answer−13\displaystyle -\frac{1}{3}

    Steps

    1. Perpendicular lines have slopes whose product is −1
    2. 3×(−13)\displaystyle 3 \times \left(-\frac{1}{3}\right)=−1\displaystyle = -1
    3. Dropping the sign would make the product +1
    4. 3×(13)\displaystyle 3 \times \left(\frac{1}{3}\right)=1\displaystyle = 1
  2. 02What is the slope of the normal line at x=4?f′(4)=3\displaystyle f'(4)=3AnswerHide
    Answer−13\displaystyle -\frac{1}{3}

    Steps

    1. Perpendicular lines have slopes whose product is −1
    2. 3×(−13)\displaystyle 3 \times \left(-\frac{1}{3}\right)=−1\displaystyle = -1
  3. 03The normal at x=3x=3. What goes in □?y=x2  ⇒  y=□(x−3)+9\displaystyle y=x^{2} \;\Rightarrow\; y = \square(x-3)+9AnswerHide
    Answer−16\displaystyle -\frac{1}{6}

    Steps

    1. First the tangent slope
    2. y′(3)\displaystyle y'(3)=6\displaystyle = 6
    3. The normal is −1 divided by the tangent slope
    4. 6×(−16)\displaystyle 6 \times \left(-\frac{1}{6}\right)=−1\displaystyle = -1
  4. 04What is the slope of the normal line at this point?f′(a)=5\displaystyle f'(a)=5AnswerHide
    Answer−15\displaystyle -\frac{1}{5}

    Steps

    1. Perpendicular lines have slopes whose product is −1
    2. 5×(−15)\displaystyle 5 \times \left(-\frac{1}{5}\right)=−1\displaystyle = -1
    3. Dropping the sign would make the product +1
    4. 5×(15)\displaystyle 5 \times \left(\frac{1}{5}\right)=1\displaystyle = 1
  5. 05What is the slope of the normal line at this point?f′(a)=4\displaystyle f'(a)=4AnswerHide
    Answer−14\displaystyle -\frac{1}{4}

    Steps

    1. Perpendicular lines have slopes whose product is −1
    2. 4×(−14)\displaystyle 4 \times \left(-\frac{1}{4}\right)=−1\displaystyle = -1
    3. Dropping the sign would make the product +1
    4. 4×(14)\displaystyle 4 \times \left(\frac{1}{4}\right)=1\displaystyle = 1
  6. 06What is the slope of the normal line at this point?f′(a)=−2\displaystyle f'(a)=-2AnswerHide
    Answer12\displaystyle \frac{1}{2}

    Steps

    1. Perpendicular lines have slopes whose product is −1
    2. −2×(12)\displaystyle -2 \times \left(\frac{1}{2}\right)=−1\displaystyle = -1
    3. Dropping the sign would make the product +1
    4. −2×(−12)\displaystyle -2 \times \left(-\frac{1}{2}\right)=1\displaystyle = 1

FAQ

Q1What is the normal line when f′(a)=0f'(a)=0?AnswerHide
A
The tangent is the horizontal line y=f(a)y=f(a), so the normal is the vertical line x=ax=a. Its slope is undefined, so the formula −1f′(a)-\frac{1}{f'(a)} does not apply.
Q2Is it enough to flip the sign, or to take the reciprocal?AnswerHide
A
You need both. If f′(a)=−2f'(a)=-2, take the reciprocal −12-\frac{1}{2} and flip its sign to get the normal slope 12\frac{1}{2}.

Published 2026-10-14 · Updated 2026-10-14