# Implicit Differentiation

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# Recall

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.

1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.
2. Sum Rule: $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$
3. Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f’(x)g(x) + f(x)g’(x)$.
4. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f’(x)g(x) - f(x)g’(x)}{(g(x))^2}$
5. Chain Rule: $\frac{d}{dx}(f(g(x))) = f’(g(x)) \cdot g’(x)$
6. $\frac{d}{dx}(e^x) = e^x, \frac{d}{dx}(\ln(x)) = \frac{1}{x}, \frac{d}{dx}(\sin(x)) = \cos(x), \frac{d}{dx}(\cos(x)) = -\sin(x), \frac{d}{dx}(\tan(x)) = \sec^2(x), \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2}.$

# Implicit Differentiation

In mathematics, some equations in x and y do not explicitly define y as a function x and cannot be easily manipulated to solve for y in terms of x, even though such a function may exist. They are defined implicitly, meaning they are expressed as equations y = f(x) rather than explicit functions of one variable. The technique of implicit differentiation allows you to find the derivative of y with respect to x without having to solve the given equation for y.

• Find the slope of the tangent line to the curve x2 + y2 = 25 at the point (3, -4).

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 $\frac{3}{4}$.

• x2 + y2 = 1 (Implicit definition).

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.

• x2 +xy + y3 = 0.

$\frac{d}{dx}(x^{2}+xy+y^3=0)⇒2x+y+x\frac{d}{dx}y+\frac{d}{dx}y^3=0 ⇒ 2x +y +x·\frac{d}{dy}y\frac{dy}{dx}+\frac{d}{dy}y^3\frac{dy}{dx}=0 ⇒ 2x +y +x·\frac{dy}{dx}+3y^2·\frac{dy}{dx}=0 ⇒ x·\frac{dy}{dx}+3y^2·\frac{dy}{dx}= -(2x+y) ⇒ \frac{dy}{dx} = -\frac{2x+y}{x+3y^2}$.

• Let’s calculate $\frac{d^2y}{dx},$ x3 + y3 = 9.

$\frac{d}{dx}(x^{3}+y^3=0)⇒3x^2+3y^2\frac{dy}{dx}=0⇒\frac{dy}{dx}=\frac{-3x^2}{3y^2}=\frac{-x^2}{y^2}$

$\frac{d^2y}{dx} = \frac{-2x·y^2-(-x^2)(2y\frac{dy}{dx})}{(y^2)^2} = \frac{-2xy^2+2x^2y\frac{dy}{dx}}{y^4} = \frac{-2xy^2+2x^2y·\frac{-x^2}{y^2}}{y^4} = \frac{-2xy^2-\frac{2x^4}{y}}{y^4} = \frac{\frac{-2xy^3-2x^4}{y}}{y^4} = \frac{-2xy^3-2x^4}{y^5} = \frac{-2x(y^3+x^3)}{y^5}$.

• Power rule for rational exponents. If u(x)=xa, then u’(x)=axa-1, where a=m/n, and m and n are both integers, e.g., y = $\sqrt{x} = x^{\frac{1}{2}} ⇒ y’ = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{\frac{-1}{2}} = \frac{1}{2\sqrt{x}}$.

y = xm/n ↔ yn = xm. Let’s apply the differential operator ddx.

$\frac{d}{dx}y^{n} = \frac{d}{dx}x^{m} = mx^{m-1}$

$\frac{d}{dx}y^{n} = (\frac{d}{dy}y^{n})\frac{dy}{dx} = ny^{n-1}\frac{dy}{dx} = mx^{m-1} ⇨ \frac{dy}{dx} = \frac{m}{n} \frac{x^{m-1}}{y^{n-1}} = \frac{m}{n} \frac{x^{m-1}}{(x^\frac{m}{n})^{n-1}} = ax^{m-1-(n-1)\frac{m}{n}}$

$m-1-(n-1)\frac{m}{n} = m -1 -m + \frac{m}{n} = \frac{m}{n} -1$

$\frac{dy}{dx} = ax^{a-1},~ where~ a=\frac{m}{n}$

• y4 + xy2 -2 = 0.

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’ = $\frac{-y^{2}}{4y^{3}+2xy}$.

# Inverse Functions

Implicit differentiation can help us solve inverse functions (you can find more detailed explanation and solved examples in this link).

• y = sin-1x ⇨ siny = x.

$\frac{d}{dx}siny=1⇨~ \frac{d}{dy}siny\frac{dy}{dx}=1⇨~ cosy\frac{dy}{dx}=1$

$y’= \frac{1}{cos(y)}=~ \frac{1}{\sqrt{1-sin^{2}y}}=~ \frac{1}{\sqrt{1-x^{2}}}$

$(sin^{-1}x)’=\frac{d}{dx}sin^{-1}x=~ \frac{1}{\sqrt{1-x^{2}}}$

• y = tan-1x ⇨ tany = x.

$\frac{d}{dx}(tany=x)⇨~ \frac{d}{dy}tany\frac{dy}{dx}=1⇨~ \frac{1}{cos^{2}y}\frac{dy}{dx}=1⇨~ y’=cos^{2}y$

$(tan^{-1}x)’=\frac{d}{dx}\tan^{-1}x=cos^{2}(y)$ [tan(y) = x ↭ $\frac{sin(y)}{cos(y)} = x ⇒ sin(y) = x·cos(y) ⇒ (x·cos(y))^2+cos^2(y) = 1 ⇒ x^2·cos^2(y) +cos^2(y) = 1 ⇒ (x^2+1)·cos^2(y) = 1 ⇒ cosy = \frac{1}{\sqrt{1+x^{2}}}$] = $\frac{1}{1+x^{2}}$

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus.
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus, College Algebra and Abstract Algebra.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
9. blackpenredpen.
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