Suppose f f f is a single-variable function, and f f f is differentiable at x 0 x_0 x 0 .
The equation of the tangent line to f f f at x 0 x_0 x 0 is:
y − f ( x 0 ) = f ′ ( x 0 ) ( x − x 0 ) y - f(x_0) = f'(x_0)(x - x_0)
y − f ( x 0 ) = f ′ ( x 0 ) ( x − x 0 )
For 2-variable functions, they will have a tangent plane. For functions with more than 2 variables, they
have a tangent space.
Suppose z = f ( x , y ) z=f(x,y) z = f ( x , y ) where f : D → R f:D\rightarrow \mathbb{R} f : D → R ; D ⊂ R 2 D\subset \mathbb{R}^2 D ⊂ R 2
and ( a , b ) ∈ D (a,b)\in D ( a , b ) ∈ D .
If f x ( a , b ) f_x(a,b) f x ( a , b ) and f y ( a , b ) f_y(a,b) f y ( a , b ) exists and Δ z \Delta{z} Δ z can be expressed in the
below form, then f f f is differentiable at ( a , b ) (a,b) ( a , b ) .
Δ z = Δ x f x ( a , b ) + Δ y f y ( a , b ) + ϵ 1 Δ x + ϵ 2 Δ y \Delta z = \Delta x f_x(a,b) + \Delta y f_y(a,b) + \epsilon_1\Delta x + \epsilon_2 \Delta y
Δ z = Δ x f x ( a , b ) + Δ y f y ( a , b ) + ϵ 1 Δ x + ϵ 2 Δ y
where ϵ 1 \epsilon_1 ϵ 1 and ϵ 2 \epsilon_2 ϵ 2 approach 0 0 0 as ( Δ x , Δ y ) (\Delta x, \Delta y) ( Δ x , Δ y )
approach ( 0 , 0 ) (0,0) ( 0 , 0 ) . OR
f f f is differentiable at ( a , b ) (a,b) ( a , b ) iff the limit exists:
lim ( x , y ) → ( a , b ) f ( x , y ) − f ( a , b ) − f x ( a , b ) Δ x − f y ( a , b ) Δ y ( x − a ) 2 + ( y − b ) 2 \lim_{(x,y)\to(a,b)} \frac{f(x,y)-f(a,b)-f_x(a,b)\Delta x -f_y(a,b)\Delta y }{\sqrt{(x-a)^2+(y-b)^2}}
( x , y ) → ( a , b ) lim ( x − a ) 2 + ( y − b ) 2 f ( x , y ) − f ( a , b ) − f x ( a , b ) Δ x − f y ( a , b ) Δ y
f f f is said to be differentiable iff it is differentiable at every point in
its domain.
If either f x f_{x} f x or f y f_y f y is non-existent at a point, then f f f is not
differentiable at that point.
If f x f_x f x and f y f_y f y are continuous throughout at an open region D D D , then
f f f is differentiable at every point of D D D .
The differentiability can also be proven by proving:
lim Δ ρ → 0 Δ z − d z Δ ρ = 0 where Δ ρ = Δ x 2 + Δ y 2 \lim_{\Delta \rho \to 0} \frac{\Delta z - \text{d}z}{\Delta \rho} = 0
\;\;\;
\text{where}
\;\;\;
\Delta \rho = \sqrt{{\Delta x}^2 + {\Delta y}^2}
Δ ρ → 0 lim Δ ρ Δ z − d z = 0 where Δ ρ = Δ x 2 + Δ y 2
f is differentiable ⟹ f is continous f \text{ is differentiable} \implies f \text{ is continous}
f is differentiable ⟹ f is continous
Suppose z = f ( x , y ) z=f(x,y) z = f ( x , y ) where f f f is a differentiable function of x x x and y y y .
d z = ∂ z ∂ x d x + ∂ z ∂ y d y = Δ x f x ( a , b ) + Δ y f y ( a , b ) \text{d}z
= \frac{\partial z}{\partial x}\,\text{d}x + \frac{\partial z}{\partial y}\,\text{d}y
=\Delta x f_x(a,b) + \Delta y f_y(a,b)
d z = ∂ x ∂ z d x + ∂ y ∂ z d y = Δ x f x ( a , b ) + Δ y f y ( a , b )
d z \text{d}z d z is called the total differential of f f f .
Let x ( t ) , y ( t ) x(t),y(t) x ( t ) , y ( t ) be single-variable, differentiable functions and f ( x , y ) f(x,y) f ( x , y ) be a
2-variable differentiable function having continuous first order partial
derivatives.
f ( x ( t ) , y ( t ) ) f\big(x(t),y(t)\big) f ( x ( t ) , y ( t ) ) is differentiable.
d f d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t \frac{\text{d}f}{\text{d}t} = \frac{\partial f}{\partial x}\frac{\text{d}x}{\text{d}t}
+ \frac{\partial f}{\partial y}\frac{\text{d}y}{\text{d}t}
d t d f = ∂ x ∂ f d t d x + ∂ y ∂ f d t d y
Can be extended for functions of more than 2 variables.