Introducing Calculus - AP Calculus AB

$f'(x) = 2x$. Using the power rule: bring down the exponent and subtract 1.

The slope of the tangent line to the function at that point. The derivative measures the steepness of the curve at any point.

$f'(x) = 5(3x^2 + 2)^4 \times 6x$. Outer function derivative is $5(3x^2 + 2)^4$, inner is $6x$.

If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.. This is the fundamental differentiation rule for polynomial terms.

$f'(x) = \text{cos}(x)$. The derivative of sine is cosine.

$f'(x) = 2x\text{sin}(x) + x^2\text{cos}(x)$. Apply product rule: $u = x^2$, $v = \sin(x)$.

$f'(x) = \frac{1}{x \text{ln}(a)}$. For logarithm base $a$, include the factor $\frac{1}{\ln(a)}$.

$f'(x) = 3x^2 - 5$. Apply power rule term by term: $3x^2 - 5 + 0$.

$f'(x) = -\text{sin}(x)$. The derivative of cosine is negative sine.

$f'(x) = 9x^2$. Apply power rule to $x^3$: $3 \cdot 3x^2 = 9x^2$.

$f'(x) = 0$. Constants have no change, so their rate of change is zero.

$f'(x) = 2x$. Using the power rule: bring down the exponent and subtract 1.

The derivative is the instantaneous rate of change of a function. This is the fundamental concept of calculus.

$f'(x) = -2x\text{e}^{-x^2}$. Chain rule: $e^{-x^2}$ times derivative of $-x^2$.

$f'(x) = -\frac{1}{|x|\text{sqrt}(x^2-1)}$. Similar to arcsec but with negative sign.

$f'(x) = -\frac{1}{\text{sqrt}(1-x^2)}$. Similar to arcsin but with negative sign.

$f'(x) = \text{sec}(x)\text{tan}(x)$. This follows from the chain rule applied to $\sec(x) = \frac{1}{\cos(x)}$.

$f'(x) = 3\text{e}^{3x}$. Chain rule: $e^{3x}$ times derivative of $3x$.

$f'(x) = 5x^4$. Using power rule: $5x^{5-1} = 5x^4$.

$f'(x) = -\frac{1}{x^2}$. Rewrite as $x^{-1}$ and apply power rule: $-1x^{-2}$.

$f'(x) = \frac{1}{|x|\text{sqrt}(x^2-1)}$. The derivative includes absolute value due to domain restrictions.

$f'(x) = \frac{1}{\text{sqrt}(1-x^2)}$. This is the standard derivative formula for inverse sine.

$f'(x) = -\frac{1}{1+x^2}$. Similar to arctan but with negative sign.

If $\frac{u}{v}$, then $\frac{vu' - uv'}{v^2}$.. For quotients: bottom times top's derivative minus top times bottom's derivative, all over bottom squared.

$f'(x) = 7$. Linear term becomes its coefficient; constant disappears.

$f'(x) = \text{sec}^2(x)$. The derivative of tangent is secant squared.

$f'(x) = \frac{3}{3x} = \frac{1}{x}$. Chain rule: derivative of $\ln(u)$ is $\frac{u'}{u}$.

$f'(x) = a^x \text{ln}(a)$. For exponential with base $a$, multiply by $\ln(a)$.

$f'(x) = 5x^4$. Using power rule: $5x^{5-1} = 5x^4$.