${dy \over du} = 7u^6$
-$7u^6 \times$
-
## Product rule for $y=uv$
$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
> the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$, must be raised, to produce that number $x$
-### Logarithmic identities
+### Logarithmic identities
+
$\log_b (xy)=\log_b x + \log_b y$
$\log_b x^n = n \log_b x$
$\log_b y^{x^n} = x^n \log_b y$
### Index identities
+
$b^{m+n}=b^m \cdot b^n$
$(b^m)^n=b^{m \cdot n}$
$(b \cdot c)^n = b^n \cdot c^n$
### Differentiating logarithms
$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
-## Solving $e^x$ etc
+## Derivative rules
| $f(x)$ | $f^\prime(x)$ |xs
| ------ | ------------- |
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
+| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ax^{nx}$ | $an \cdot e^{nx}$ |
| $\log_e {ax}$ | $1 \over x$ |
| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
+| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
+| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
+Reciprocal derivatives:
+
+$${{dy \over dx} \over 1} = dx \over dy$$
+
+## Differentiating $x=f(y)$
+
+Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}}$.
+
+$${dy \over dx} = {1 \over {dx \over dy}}$$
+
+## Second derivative
+
+$$f(x) \implies f^\prime (x) \implies f^{\prime\prime}(x)$$
+
+$$\therefore y \implies {dy \over dx} \implies {d({dy \over dx}) \over dx} \implies {d^2 y \over dx^2}$$
+
+Order of polynomial $n$th derivative decrements each time the derivative is taken
+
+### Maxima and minima
+
+- if $f^\prime (a) = 0$ and $f^{\prime\prime}(a) > 0$, then point $(a, f(a))$ is a local min (curve is concave up)
+
+- if $f^\prime (a) = 0$ and $f^{\prime\prime} (a) < 0$, then point $(a, f(a))$ is local max (curve is concave down)
+- if $f^{\prime\prime}(a) = 0$, then point $(a, f(a))$ is a point of inflection
+- - if also $f^\prime(a)=0$, then it is a stationary point of inflection
+
+*Point of inflection* - point of maximum gradient (either +ve or -ve)
+
## Antidifferentiation
$$y={x^{n+1} \over n+1} + c$$
To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
-## Kinematics
+## Rates
+
+### Related rates
+
+$${da \over db} \quad \text{change in } a \text{ with respect to } b$$
-$${dV \over dt} = {\operatorname{change in volume} \over \operatorname{respect to time}}$$
+#### Gradient at a point on parametric curve
-**position $x$** - distance from origin or fixed point
-**displacement $s$** - change in position from starting point (vector)
-**velocity $v$** - change in position with respect to time
-**acceleration $a$** - change in velocity
-**speed** - magnitude of velocity
+$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
-$$v_{\operatorname{avg}}={\Delta x \over \Delta t}={{x_2 - x_1} \over {t_2 - t_1}}$$
-$$\operatorname{speed}_{\operatorname{avg}}={\Delta v \over \Delta t}$$
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$