The real problem of humanity is the following: We have Paleolithic emotions, medieval institutions and godlike technology. And it is terrifically dangerous, and it is now approaching a point of crisis overall, Edward O. Wilson.
The derivative of a function at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change, the ratio of the instantaneous change in the dependent variable to that of the independent variable.
Definition. A function f(x) is differentiable at a point “a” of its domain, if its domain contains an open interval containing “a”, and the limit $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$ exists, f’(a) = L = $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then |L-$\frac {f(a+h)-f(a)}{h}$|< ε.
In mathematics, some equations involving two variables, x and y, do not explicitly define y as a function of x. This means that even though y can be expressed as a function of x implicitly, the equation might be too complex or impossible to rearrange in a way that solves for y explicitly in terms of x. These equations are said to be defined implicitly.
For example, consider an equation like x2 + y2 = 1. This equation implicitly defines a relationship between x and y, but it does not express y directly as a function of x (such as y = f(x)). However, there might still exist a function y=f(x) that satisfies this equation for each x.
The technique of implicit differentiation allows us to find the derivative of y with respect to x (denoted as $\frac{dy}{dx}$) without needing to solve the equation explicitly for y.
2x +2y·y’ = 0 ⇒ 2y·y’ = -2x. y·y’ = -x ⇒ y’ = $\frac{-x}{y}$ ⇒ y’(3) = $\frac{-x}{y} = \frac{-3}{-4} = \frac{3}{4}$. Therefore, the slope of the tangent line at (3, -4) is $\frac{3}{4}$.
y2 = 1 - x2 ⇨ $y=\pm\sqrt{1-x^{2}}$ (Explicit definition). Let’s take the positive branch, $y=+\sqrt{1-x^{2}}=(1-x^{2})^{\frac{1}{2}}$
y’ = $\frac{1}{2}(1-x^{2})^{\frac{-1}{2}}(-2x) = -x(1-x^{2})^{\frac{-1}{2}}=\frac{-x}{\sqrt{1-x^{2}}}$
Or alternatively, x2 + y2 = 1 ⇨ $\frac{d}{dx}(x^{2}+y^{2}=1)~⇨2x+2yy’=0⇨~y’=\frac{-x}{y}.$ The implicit way does not need to take only one branch.
$\frac{d}{dx}(x^{2}+xy+y^3=0)⇒2x+y+x\frac{d}{dx}y+\frac{d}{dx}y^3=0 ⇒ 2x +y +x·\frac{d}{dy}y\frac{dy}{dx}+\frac{d}{dy}y^3\frac{dy}{dx}=0 ⇒ 2x +y +x·\frac{dy}{dx}+3y^2·\frac{dy}{dx}=0 ⇒ x·\frac{dy}{dx}+3y^2·\frac{dy}{dx}= -(2x+y) ⇒ \frac{dy}{dx} = -\frac{2x+y}{x+3y^2}$.
$\frac{d}{dx}(x^{3}+y^3=0)⇒3x^2+3y^2\frac{dy}{dx}=0⇒\frac{dy}{dx}=\frac{-3x^2}{3y^2}=\frac{-x^2}{y^2}$
$\frac{d^2y}{dx} = \frac{-2x·y^2-(-x^2)(2y\frac{dy}{dx})}{(y^2)^2} = \frac{-2xy^2+2x^2y\frac{dy}{dx}}{y^4} = \frac{-2xy^2+2x^2y·\frac{-x^2}{y^2}}{y^4} = \frac{-2xy^2-\frac{2x^4}{y}}{y^4} = \frac{\frac{-2xy^3-2x^4}{y}}{y^4} = \frac{-2xy^3-2x^4}{y^5} = \frac{-2x(y^3+x^3)}{y^5}$.
y = xm/n ↔ yn = xm. Let’s apply the differential operator d⁄dx.
$\frac{d}{dx}y^{n} = \frac{d}{dx}x^{m} = mx^{m-1}$
$\frac{d}{dx}y^{n} = (\frac{d}{dy}y^{n})\frac{dy}{dx} = ny^{n-1}\frac{dy}{dx} = mx^{m-1} ⇨ \frac{dy}{dx} = \frac{m}{n} \frac{x^{m-1}}{y^{n-1}} = \frac{m}{n} \frac{x^{m-1}}{(x^\frac{m}{n})^{n-1}} = ax^{m-1-(n-1)\frac{m}{n}}$
$m-1-(n-1)\frac{m}{n} = m -1 -m + \frac{m}{n} = \frac{m}{n} -1 $
$\frac{dy}{dx} = ax^{a-1},~ where~ a=\frac{m}{n}$
Explicit. $y^{2}=\frac{-x\pm\sqrt{x^{2}+8}}{2}, y=\pm\sqrt{\frac{-x\pm\sqrt{x^{2}+8}}{2}}$. It is not a good approach.
Implicit. 4y3y’+ y2 + 2yy’x = 0. ⇨ (4y3+2xy)y’ + y2 = 0 ⇨ y’ = $\frac{-y^{2}}{4y^{3}+2xy}$.
Implicit differentiation can help us solve inverse functions (you can find more detailed explanation and solved examples in this link).
Now, differentiate both sides with respect to x: $\frac{d}{dx}sin(y)=1⇨~ cos(y)\frac{dy}{dx}=1⇨~\frac{dy}{dx} = \frac{1}{cos(y)}$
$y’= \frac{1}{cos(y)}=~ \frac{1}{\sqrt{1-sin^{2}y}}=~ \frac{1}{\sqrt{1-x^{2}}}$
$\frac{d}{dx}(tany=x)⇨~ \frac{d}{dy}tany\frac{dy}{dx}=1⇨~ \frac{1}{cos^{2}y}\frac{dy}{dx}=1⇨~ y’=cos^{2}y$
$(tan^{-1}x)’=\frac{d}{dx}\tan^{-1}x=cos^{2}(y)$ [tan(y) = x ↭ $ \frac{sin(y)}{cos(y)} = x ⇒ sin(y) = x·cos(y) ⇒ (x·cos(y))^2+cos^2(y) = 1 ⇒ x^2·cos^2(y) +cos^2(y) = 1 ⇒ (x^2+1)·cos^2(y) = 1 ⇒ cosy = \frac{1}{\sqrt{1+x^{2}}}$] = $\frac{1}{1+x^{2}}$