Composite, Implicit, and Inverse Functions

$\int_a^b |f(x)| ,dx$

$\int_a^b f'(x) ,dx$

$\sum_{x=a}^{b} f(x)\Delta x$

$\int_a^b f(x) ,dx$

$e^{2x}$

$2e^{2x + 3}$

$e^{2x + 3}$

$2e^{2x}$

5e^x + 2/x

5e^x - 1/x

5e^x + 1/x

5e^x - 2/x

$2\cos(x) - 3\cos(x)$

$2\cos(x) + 3\sin(x)$

$2\cos(x) - 3\sin(x)$

$2\sin(x) + 3\sin(x)$

Apply l’Hôpital’s rule directly on the original limit expression without using Stirling's approximation

Apply Stirling’s approximation and evaluate as $n$ approaches infinity yielding $e$

Utilize Taylor series expansion equation governing factorial growth before taking limits

Consider $\frac{n!}{n^n}$ under the $n$th root after utilizing Stirling’s approximation leading toward infinity

$4(2x^2 + 3)$

$8(2x^2 + 3)$

$4x(2x^2 + 3)$

$8x(2x^2 + 3)$

Chain Rule

Implicit Differentiation

Power Rule

Product Rule

$\pi$

$\sin(\pi)$

Zero.

-$\pi$

There exists some $c$ in $(a,b)$ such that $f(c)=0$ if $f$ is only continuous, not necessarily differentiable on $(a,b)$.

There are multiple points $c$ in $[a,b]$ where $f'(c)=0$ as long as $f$ has no discontinuities on $[a,b]$.

Rolle's theorem guarantees no such connection between continuity and differentiability over any given interval.

There exists some $c$ in $(a,b)$ such that $f'(c)=0$ if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

$8(2x - 1)^3$

$16x^3 - 12x^2 + 6x - 1$

$16x^3 - 12x^2 + 6x$

$8(2x - 1)^3 - 1$