Tangent line using linearity rules for differentiation

Differentiate term by term using the sum/difference rule and constant multiple rule, applying the power rule $\frac{d}{dx}[x^n] = nx^{n-1}$ to each term: $f'(x) = -36x^3 + 6x^2 - 12x + 1.$ This gives the general slope function for any $x$.

Evaluate at $x=1$: $f(1) = -9 + 2 - 6 + 1 - 3 = -15$, so the point of tangency is $(1, -15)$. The slope is $f'(1) = -36 + 6 - 12 + 1 = -41$. Using point-slope form, $y - (-15) = -41(x - 1)$, which simplifies to $y = -41x + 26$.

The derivative of the original function is negative thirty six x cubed plus six x squared minus twelve x plus one. The term containing x cubed comes from differentiating negative nine x to the fourth power using the power rule, giving negative thirty six x cubed. So the coefficient of that term is negative thirty six.