# If y=ln(f(x)), dy/dx=(f'(x))/f(x)

Prove that if:

$y=\ln { \left( f\left( x \right) \right) } ,\quad \frac { dy }{ dx } =\frac { f$

Firstly, say that:

$y=h\left( x \right) \\ \\ g=f\left( x \right)$

So:

# Vector Magnitude Proof

Want to find the magnitude of a vector?

You quite simply have to know that:

${ \left| \underline { v } \right| }^{ 2 }={ \underline { v } }^{ 2 }$

And that if:

$\underline { v } =\left( \begin{matrix} x \\ y \\ z \end{matrix} \right) ,\quad { \underline { v } }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$

So:

$If\quad { \underline { v } }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 },\\ \\ \therefore \quad { \left| \underline { v } \right| }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \therefore \quad \left| \underline { v } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \\$

# Vector Proof – Angle Between Two Vectors

Prove that:

$cos\theta =\frac { \underline { a } \cdot \underline { b } }{ \left| \underline { a } \right| \left| \underline { b } \right| }$

Firstly, look at the image below:

Also know that:

$cosC=\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab }$

$\left| \underline { a } \right| \left| \underline { a } \right| ={ \underline { a } }^{ 2 }\\ \\ \left| \underline { b } \right| \left| \underline { b } \right| ={ \underline { b } }^{ 2 }\\ \\ \left| \underline { b } -\underline { a } \right| \left| \underline { b } -\underline { a } \right| ={ \left( \underline { b } -\underline { a } \right) }^{ 2 }$

From the image, you’ll be able to see that:

$a=\left| \underline { b } \right| \\ \\ b=\left| \underline { a } \right| \\ \\ c=\left| \underline { b } -\underline { a } \right| \\ \\ cosC=cos\theta ,\quad \therefore \quad C=\theta$

Now:

# If y=secx, dy/dx=secxtanx

Prove that:

$y=secx,\quad \frac { dy }{ dx } =secxtanx$

Proof:

# If y=tanx, dy/dx=(secx)^2

Prove that if:

$y=tanx,\quad \frac { dy }{ dx } ={ sec }^{ 2 }x$

Proof:

Prove that if:

$y=tankx,\quad \frac { dy }{ dx } =k{ sec }^{ 2 }kx$

Proof:

$y=tankx=tanu\\ \\ \frac { dy }{ du } ={ sec }^{ 2 }u,\\ \\ u=kx,\quad \frac { du }{ dx } =k\\ \\ \frac { du }{ dx } \cdot \frac { dy }{ du } =k{ sec }^{ 2 }kx\\ \\ \therefore \quad \frac { dy }{ dx } =k{ sec }^{ 2 }kx$

# Vector Proof (1)

Prove that:

$\left( \begin{matrix} { a }_{ 1 } \\ { a }_{ 2 } \\ { a }_{ 3 } \end{matrix} \right) \left( \begin{matrix} { b }_{ 1 } \\ { b }_{ 2 } \\ { b }_{ 3 } \end{matrix} \right) ={ a }_{ 1 }{ b }_{ 1 }+{ a }_{ 2 }{ b }_{ 2 }+{ a }_{ 3 }{ b }_{ 3 }\\ \\$

Firstly, look at the image below.

You should know that, if $\underline { a } \\ \\$ and $\underline { b } \\ \\$ are perpendicular $\underline { a } \cdot \underline { b } =0\\ \\$.

You should also know these rules:

$\left| \underline { i } \right| \left| \underline { i } \right| ={ \underline { i } }^{ 2 }=1\cdot 1=1\\ \\ \left| \underline { j } \right| \left| \underline { j } \right| ={ \underline { j } }^{ 2 }=1\cdot 1=1\\ \\ \left| \underline { k } \right| \left| \underline { k } \right| ={ \underline { k } }^{ 2 }=1\cdot 1=1\\ \\$

Knowing these rules, we can say that:

*Click on the proof above to see it in full.

# Cosine Rule Transformation

Don’t waste time memorising two sets of formulas that are simply the same.

# Sine Rule Derivation

Use the formulas you’d use to calculate the area of a triangle. See the magic emerge.

$\frac { 1 }{ 2 } bcsinA=\frac { 1 }{ 2 } acsinB\\ \\ bcsinA=acsinB\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ b } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } \cdot \frac { 1 }{ a } =\frac { asinB }{ b } \cdot \frac { 1 }{ a } \\ \\ \frac { sinA }{ a } =\frac { sinB }{ b } \\ \\ \\ OR:\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ sinB } =\frac { asinB }{ sinB } \\ \\ \frac { bsinA }{ sinB } =\frac { a }{ 1 } \\ \\ \frac { bsinA }{ sinB } \cdot \frac { 1 }{ sinA } =\frac { a }{ 1 } \cdot \frac { 1 }{ sinA } \\ \\ \frac { b }{ sinB } =\frac { a }{ sinA }$

# Trigonometric Proofs

Rules:

$sec\theta =\frac { 1 }{ cos\theta } \quad \therefore \quad { sec }^{ 2 }\theta =\frac { 1 }{ { cos }^{ 2 }\theta } \\ \\ cosec\theta =\frac { 1 }{ sin\theta } \quad \therefore \quad { cosec }^{ 2 }\theta =\frac { 1 }{ { sin }^{ 2 }\theta } \\ \\ cot\theta =\frac { 1 }{ tan\theta } =\frac { 1 }{ \frac { sin\theta }{ cos\theta } } =\frac { cos\theta }{ sin\theta } \\ \\ \therefore \quad { cot }^{ 2 }\theta =\frac { 1 }{ { tan }^{ 2 }\theta } =\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta }$

Transformation 1:

${ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1\\ \\ \frac { { sin }^{ 2 }\theta }{ { sin }^{ 2 }\theta } +\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta } \\ \\ 1+{ cot }^{ 2 }\theta ={ cosec }^{ 2 }\theta$

Transformation 2:

${ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1\\ \\ \frac { { sin }^{ 2 }\theta }{ { cos }^{ 2 }\theta } +\frac { { cos }^{ 2 }\theta }{ { cos }^{ 2 }\theta } =\frac { 1 }{ { cos }^{ 2 }\theta } \\ \\ { tan }^{ 2 }\theta +1={ sec }^{ 2 }\theta$