# How to differentiate y=arctanx

Below I’m going to demonstrate how to integrate y=arctanx…

Firstly, we need to know that: $\sin ^{ 2 }{ y } +\cos ^{ 2 }{ y } =1\\ \\ \frac { \sin ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } +\frac { \cos ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } =\frac { 1 }{ \cos ^{ 2 }{ y } } \\ \\ \tan ^{ 2 }{ y } +1=\sec ^{ 2 }{ y }$

We also need to know that: $x=\tan { y } \\ \\ x=\frac { \sin { y } }{ \cos { y } } \\ \\ x\cdot \cos { y } =\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } +x\cdot \left( -\sin { y } \right) =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } -x\sin { y } =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +x\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +\frac { \sin ^{ 2 }{ y } }{ \cos { y } } \\ \\ \frac { 1 }{ \cos { y } } \cdot \frac { dx }{ dy } \cdot \cos { y } =\frac { 1 }{ \cos { y } } \left( \cos { y+\frac { \sin ^{ 2 }{ y } }{ \cos { y } } } \right) \\ \\ \frac { dx }{ dy } =1+\tan ^{ 2 }{ y } \\ \\ \therefore \quad \frac { dx }{ dy } =\sec ^{ 2 }{ y }$

And finally: $\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Now, using implicit differentiation: $y=\arctan { x } \\ \\ \tan { y } =x\\ \\ \sec ^{ 2 }{ y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \sec ^{ 2 }{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \tan ^{ 2 }{ y+1 } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ { x }^{ 2 }+1 }$

# Logarithmic Differentiation As Seen In Video…

In this video, Patrick JMT demonstrated how to do logarithmic differentiation.

Now I’m going to show you how I’d solve the same problem…

Firstly we must know that: $y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ If\quad y=u\cdot v,\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ u=x,\quad \frac { du }{ dx } =1\\ \\ v=\ln { \left( \ln { x } \right) } =\ln { q } \\ \\ \frac { dv }{ dq } =\frac { 1 }{ q } =\frac { 1 }{ \ln { x } } \\ \\ q=\ln { x } ,\quad \frac { dq }{ dx } =\frac { 1 }{ x } \\ \\ \therefore \quad \frac { dv }{ dx } =\frac { 1 }{ x\ln { x } } \\ \\ \therefore \quad \frac { dy }{ dx } =x\cdot \frac { 1 }{ x\ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) }$

Now: $y={ \left( \ln { x } \right) }^{ x }\\ \\ \ln { y=\ln { \left( { \left( \ln { x } \right) }^{ x } \right) } } \\ \\ \ln { y } =x\ln { \left( \ln { x } \right) } \\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\} \\ \\ \frac { dy }{ dx } ={ \left( \ln { x } \right) }^{ x }\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\}$

# Derivative – 27/02/15 $n\ln { x } =y\\ \\ \ln { \left( { x }^{ n } \right) } =y\\ \\ \log _{ e }{ \left( { x }^{ n } \right) } =y\\ \\ { e }^{ y }={ x }^{ n }\\ \\ { e }^{ y }\frac { dy }{ dx } =n{ x }^{ n-1 }\\ \\ \frac { 1 }{ { e }^{ y } } \cdot { e }^{ y }\frac { dy }{ dx } =n{ x }^{ n-1 }\cdot \frac { 1 }{ { e }^{ y } } \\ \\ \frac { dy }{ dx } =n{ x }^{ n-1 }\cdot { x }^{ -n }\\ \\ \frac { dy }{ dx } =n{ x }^{ n-1+\left( -n \right) }\\ \\ \frac { dy }{ dx } =n{ x }^{ -1 }\\ \\ \frac { dy }{ dx } =\frac { n }{ x }$

# Anti-Derivative Proof – 27/02/15 $y=\frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \ln { y } =\ln { \left( \frac { { x }^{ \left( n+1 \right) } }{ \left( n+1 \right) } \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } -\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } +{ C }_{ 2 }\\ \\ \therefore \quad { C }_{ 2 }=-\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\left( n+1 \right) \cdot \ln { x } +{ C }_{ 2 }\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \cdot y\\ \\ \frac { dy }{ dx } =\left( n+1 \right) \cdot { x }^{ -1 }\cdot \frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1 }\cdot { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1+\left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ n }\\ \\ \therefore \quad \int { { x }^{ n } } dx=\frac { 1 }{ \left( n+1 \right) } \cdot { x }^{ n+1 }+C$

# 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: 