Exercise 213Calculus › Single-variable calculus › Differentiation › Differentiable functionsHigh schoolAdded 2 Nov 2026

Derivative and tangent line of a cubic

#derivative#tangent line#polynomials

Statement

Let $f(x)=5x^3+7x^2-4x+1$ . Compute $f'(x)$ and find the equation of the tangent line to the graph of $f$ at the point with abscissa $a=2$ .

Hint

Use $(x^n)'=nx^{n-1}$ and the tangent line equation $y=f'(a)(x-a)+f(a)$ .

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Solution

Derivative. Differentiating term by term, $f'(x)=15x^2+14x-4$ .

At $a=2$ .
$f'(2)=60+28-4=84,\qquad f(2)=40+28-8+1=61.$

Tangent line. $y=84(x-2)+61$ , that is
$y=84x-107.$

Meaning. Near $x=2$ the curve is almost indistinguishable from this line: $f(2+h)\approx61+84h$ for small $h$ . Derivatives are, above all, slopes of best linear approximations.

Derivative and tangent line of a cubic

Statement

Solution

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