Here it is, the rule that replaces the limit definition for any power of \(x\): bring the exponent down in front as a coefficient, then subtract one from the exponent. That's it. Slide the exponent and watch the derivative \(f'\) (gold, dashed) track the slope of \(f\) (lime) everywhere.
What you're looking at: \(f(x)=x^n\) in lime and its derivative \(f'(x)\) in gold. Notice \(f'\) is always one degree lower than \(f\).
The boundary: the variable must be the base, not the exponent. The power rule differentiates \(x^n\) — it does not apply to \(2^x\), where \(x\) is up in the exponent.
The real power of the power rule: \(n\) can be any real number. Roots and reciprocals just need to be rewritten as exponents first — then the same “down and minus one” move handles them. Tap each.
The power rule only sees \(x^r\). It can't read a fraction bar or a root symbol — so convert to a single exponent, differentiate, then convert back if the problem wants it pretty.
Three rules so natural they barely need names: the derivative of a constant is zero, constants factor straight out, and the derivative of a sum is the sum of the derivatives. Together they mean you can attack a polynomial one term at a time. Step through it.
Each term is independent: the \(x^4\) term doesn't care about the \(x^2\) term. The constant \(-1\) contributes nothing — its derivative is \(0\). This is why polynomial derivatives are so fast.
Zoom in on a smooth curve at a point and it looks like a straight line — you saw that with the magnifier last lesson. That line is the tangent: the one straight line that best matches the curve right at that point. Of all the lines you could draw through the point, the tangent is the one that hugs the curve most closely.
What you're looking at: a line through the point on \(f(x)=x^2\). The shaded band is the gap between line and curve. Tilt away and the gap grows; the tangent makes it smallest.
This is what “the slope at a point” really means: the tangent is the line the curve is heading along right there. Its slope is exactly \(f'(a)\) — the derivative.
So a tangent line needs just two ingredients, and you already have both: a point on the curve, \(\big(a, f(a)\big)\), and a slope at that point, \(f'(a)\). Drop them into point-slope form and you have the equation. Step through it:
What you're looking at: pick where on the curve you want the tangent, then build its equation one ingredient at a time — point, slope, line.
The formula: \(y = f(a) + f'(a)\,(x-a)\). Read it as “start at the point's height, then rise at the tangent's slope.” This is the AP exam's most-asked tangent question, every year.
Now the rule earns its keep. A horizontal tangent is a point where the slope is zero — a peak, a valley, a flat spot. Since the derivative is the slope, you find them by solving \(f'(x)=0\). Drag the point along \(f(x)=x^3-3x\) and watch the tangent flatten.
What you're looking at: the tangent turns lime and level exactly where \(f'(x)=0\) — at \(x=-1\) and \(x=1\) for this curve.
To find them by hand: set \(f'(x)=3x^2-3=0\), so \(x^2=1\), giving \(x=\pm1\). Horizontal tangents are just the zeros of the derivative.
The normal line at a point is the line perpendicular to the tangent there. Perpendicular slopes are negative reciprocals, so if the tangent slope is \(m\), the normal slope is \(-\tfrac1m\). One derivative gives you both lines.
What you're looking at: the gold tangent and the blue normal, always at right angles, on \(f(x)=0.6x^2\).
To write the normal line's equation, use the point and the slope \(-\tfrac1m\) in point-slope form. The only place this breaks is where \(m=0\) — a horizontal tangent has a vertical normal.
The whole power rule in four words: exponent down, minus one. \(\frac{d}{dx}x^n=nx^{n-1}\).
Roots and reciprocals become exponents: \(\sqrt{x}=x^{1/2}\), \(\tfrac1{x^n}=x^{-n}\). Then the rule applies as usual.
Differentiation is linear: constants factor out, sums split apart. A polynomial differentiates one term at a time.
A horizontal tangent is where the slope is zero. Find them by solving \(f'(x)=0\) — the zeros of the derivative.
Five problems — the power rule on whole, negative, and fractional powers, a full polynomial, and a horizontal-tangent hunt. Try each before revealing.
Eight fast checks across the whole lesson.
\(\frac{d}{dx}x^3\) is \(3x^2\), not \(3x^3\). Bring the exponent down and reduce it. Both steps, every time.
\(\frac{d}{dx}(7)=0\), but \(\frac{d}{dx}(7x)=7\). A lone constant vanishes; a constant times \(x\) leaves the constant behind.
You can't power-rule a fraction bar or a root symbol directly. Convert \(\frac{1}{x^2}\) to \(x^{-2}\) and \(\sqrt{x}\) to \(x^{1/2}\) first.
The power rule is for a variable base with a constant exponent. When the variable is in the exponent (\(2^x\), \(e^x\)), it does not apply — that's a different rule, coming in the next lesson.
\(\frac{d}{dx}x^{-2}=-2x^{-3}\). The new exponent is \(-3\), not \(-1\) — subtracting one from \(-2\) goes down to \(-3\).
Horizontal tangents come from \(f'(x)=0\), not \(f(x)=0\). One is where the slope is flat; the other is where the curve crosses the axis.