# Integration Proof (1)

Prove that:

$\int { { x }^{ n }dx=\frac { { x }^{ n+1 } }{ n+1 } } +C$

Proof:

# Implicit Differentiation Rules

Rule 1:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Rule 2:

# 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:

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