1. Derivative using Definition (First Principle)
વ્યાખ્યાની મદદથી વિકલન (પ્રથમ સિદ્ધાંત)
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$
⚠️ Need a quick review? (પુનરાવર્તનની જરૂર છે?)
Derivatives use a lot of Trigonometry and Limit rules. Brush up on your basics before solving these!
વિકલનમાં ત્રિકોણમિતિ અને લક્ષ ના નિયમો મોટા પ્રમાણમાં વપરાય છે. પહેલા પાયાનું જ્ઞાન તાજું કરો!
Practice Basics ↺
Example 1: Find the derivative of $f(x) = x^3$ using definition.
વ્યાખ્યાની મદદથી $x^3$ નું $x$ સાપેક્ષ વિકલિત શોધો.
Step 1 (Substitute into formula): $f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$
Step 2 (Expand using binomial theorem): $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$. Substitute this back:
$f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}$
Step 3 (Simplify numerator): Cancel $x^3 - x^3 = 0$.
$f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$
Step 4 (Factor out h): $f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$
Step 5 (Apply Limit): Cancel $h$ and substitute $h=0$.
$f'(x) = 3x^2 + 3x(0) + (0)^2 = 3x^2$
Example 2: Find the derivative of $f(x) = \sqrt{x}$ using definition.
વ્યાખ્યાની મદદથી $\sqrt{x}$ નું વિકલન મેળવો.
Step 1 (Formula): $f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$
Step 2 (Rationalize): Multiply numerator and denominator by the conjugate $(\sqrt{x+h} + \sqrt{x})$.
$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$
Step 3 (Apply difference of squares): $(a-b)(a+b) = a^2 - b^2$.
$f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$
Step 4 (Simplify & Limit): $\lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$
Substitute $h=0 \implies f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$
Example 3: Find the derivative of $f(x) = \cos x$ using definition.
વ્યાખ્યાની મદદથી $y = \cos x$ નું વિકલન શોધો.
Step 1 (Formula): $f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$
Step 2 (Trig Identity): Use $\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$. Let $C = x+h$, $D = x$.
$C+D = 2x+h$, $C-D = h$.
Step 3 (Substitute): $f'(x) = \lim_{h \to 0} \frac{-2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}$
Step 4 (Rearrange for standard limit): Rewrite as $\lim_{h \to 0} \left[ -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2} \right]$
Step 5 (Apply Limit): Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, as $h \to 0$, $h/2 \to 0$.
$f'(x) = -\sin(x + 0) \cdot 1 = -\sin x$
2. Working Rules & Chain Rule
કાર્યકારી નિયમો અને સાંકળ નિયમ
Chain Rule: Used for composite functions. If $y$ is a function of $u$, and $u$ is a function of $x$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
સાંકળ નિયમ: સંયોજિત વિધેયો માટે. જો $y$ એ $u$ નું વિધેય હોય અને $u$ એ $x$ નું વિધેય હોય, તો $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
Example 1 (Chain Rule)
If $y = \log(x + \sqrt{x^2 + a^2})$, then find $\frac{dy}{dx}$.
જો $y = \log(x + \sqrt{x^2 + a^2})$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Outer function): Derivative of $\log(u)$ is $\frac{1}{u}$.
$\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + a^2})$
Step 2 (Inner function): Differentiate the terms inside. $\frac{d}{dx}(x) = 1$. Using chain rule for the square root: $\frac{d}{dx}((x^2+a^2)^{1/2}) = \frac{1}{2\sqrt{x^2+a^2}} \cdot 2x = \frac{x}{\sqrt{x^2+a^2}}$.
Step 3 (Combine): $\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left[1 + \frac{x}{\sqrt{x^2 + a^2}}\right]$
Step 4 (Simplify): Take LCM inside the bracket: $\frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}$.
$\frac{dy}{dx} = \frac{1}{(x + \sqrt{x^2 + a^2})} \cdot \frac{(x + \sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}}$
Cancel the common term. Final Answer: $\frac{1}{\sqrt{x^2 + a^2}}$
Example 2 (Chain Rule)
If $y = \log(\sec x + \tan x)$ then find $\frac{dy}{dx}$.
જો $y = \log(\sec x + \tan x)$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Outer function): $\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x)$
Step 2 (Inner functions): $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\tan x) = \sec^2 x$.
$\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)$
Step 3 (Factor & Cancel): Take $\sec x$ common from the numerator: $\sec x(\tan x + \sec x)$.
$\frac{dy}{dx} = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \sec x$
Example 3 (Sum & Chain Rules)
If $y = e^{\tan x} + \log(\sin x)$ then find $\frac{dy}{dx}$.
જો $y = e^{\tan x} + \log(\sin x)$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Apply Sum Rule): $\frac{dy}{dx} = \frac{d}{dx}(e^{\tan x}) + \frac{d}{dx}(\log(\sin x))$
Step 2 (Chain Rule on term 1): $\frac{d}{dx}(e^{\tan x}) = e^{\tan x} \cdot \frac{d}{dx}(\tan x) = e^{\tan x} \sec^2 x$
Step 3 (Chain Rule on term 2): $\frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{\cos x}{\sin x} = \cot x$
Step 4 (Combine): Final Answer: $\frac{dy}{dx} = e^{\tan x} \sec^2 x + \cot x$
3. Parametric & Logarithmic Differentiation
પ્રચલ અને લઘુગણકીય વિધેયોનું વિકલન
Parametric (પ્રચલ): When $x$ and $y$ are given in terms of a third variable $t$ or $\theta$. Rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Example 1 (Parametric)
If $x = a(t + \sin t)$ and $y = a(1 - \cos t)$, find $\frac{dy}{dx}$.
જો $x = a(t + \sin t)$ અને $y = a(1 - \cos t)$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Find dx/dt): Differentiate $x$ w.r.t $t$. $\frac{dx}{dt} = a(1 + \cos t)$
Step 2 (Find dy/dt): Differentiate $y$ w.r.t $t$. $\frac{dy}{dt} = a(0 - (-\sin t)) = a \sin t$
Step 3 (Divide): $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t}$
Step 4 (Trig Identity Simplification): Use $\sin t = 2 \sin(t/2) \cos(t/2)$ and $1 + \cos t = 2 \cos^2(t/2)$.
$\frac{dy}{dx} = \frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)} = \frac{\sin(t/2)}{\cos(t/2)} = \tan\left(\frac{t}{2}\right)$
Example 2 (Parametric)
If $x = 2e^{-t}$ and $y = \frac{1}{2}e^t$, find $\frac{dy}{dx}$.
જો $x = 2e^{-t}$ અને $y = \frac{1}{2}e^t$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Find dx/dt): $\frac{dx}{dt} = 2 \cdot (-1) \cdot e^{-t} = -2e^{-t}$
Step 2 (Find dy/dt): $\frac{dy}{dt} = \frac{1}{2}e^t$
Step 3 (Divide): $\frac{dy}{dx} = \frac{\frac{1}{2}e^t}{-2e^{-t}}$
Step 4 (Simplify): Move $e^{-t}$ to the numerator: $\frac{dy}{dx} = -\frac{1}{4}e^t \cdot e^t = -\frac{1}{4}e^{2t}$
Logarithmic (લઘુગણકીય): Used when a variable is in the power of another variable (e.g., $y = [f(x)]^{g(x)}$).
Example 3 (Logarithmic)
If $y = x^x$ then find $\frac{dy}{dx}$.
જો $y = x^x$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Take Log): Apply natural logarithm to both sides: $\ln y = \ln(x^x) \implies \ln y = x \ln x$
Step 2 (Differentiate implicitly): Differentiate w.r.t $x$. Left side becomes $\frac{1}{y} \frac{dy}{dx}$. Right side needs Product Rule ($uv' + vu'$).
Step 3 (Product Rule): $\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(x) = x\left(\frac{1}{x}\right) + \ln x(1) = 1 + \ln x$
Step 4 (Isolate dy/dx): Multiply both sides by $y$. $\frac{dy}{dx} = y(1 + \ln x)$.
Replace $y$ with original function: $\frac{dy}{dx} = x^x(1 + \ln x)$
Example 4 (Logarithmic)
If $y = (\sin x)^{\tan x}$ then find $\frac{dy}{dx}$.
જો $y = (\sin x)^{\tan x}$ હોય તો $\frac{dy}{dx}$ શોધો.
Step 1 (Take Log): $\ln y = \tan x \cdot \ln(\sin x)$
Step 2 (Differentiate implicitly): $\frac{1}{y}\frac{dy}{dx} = \tan x \cdot \frac{d}{dx}(\ln(\sin x)) + \ln(\sin x) \cdot \frac{d}{dx}(\tan x)$
Step 3 (Chain Rule inner terms): $\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
$\frac{1}{y}\frac{dy}{dx} = \tan x \cdot \cot x + \ln(\sin x) \cdot \sec^2 x$
Step 4 (Simplify): Since $\tan x \cdot \cot x = 1$, we get $\frac{1}{y}\frac{dy}{dx} = 1 + \sec^2 x \ln(\sin x)$.
Final Answer: $\frac{dy}{dx} = (\sin x)^{\tan x} [1 + \sec^2 x \ln(\sin x)]$
4. Successive Differentiation ("Prove That")
ઉત્તરોત્તર વિકલન
Example 1
If $y = A \cos pt + B \sin pt$, prove that $\frac{d^2y}{dt^2} + p^2y = 0$.
જો $y = A \cos pt + B \sin pt$ હોય તો સાબિત કરો કે $\frac{d^2y}{dt^2} + p^2y = 0$.
Step 1 (First Derivative): Differentiate w.r.t $t$.
$\frac{dy}{dt} = A(-\sin pt \cdot p) + B(\cos pt \cdot p) = -Ap \sin pt + Bp \cos pt$
Step 2 (Second Derivative): Differentiate again w.r.t $t$.
$\frac{d^2y}{dt^2} = -Ap(\cos pt \cdot p) + Bp(-\sin pt \cdot p) = -Ap^2 \cos pt - Bp^2 \sin pt$
Step 3 (Factorize): Factor out $-p^2$.
$\frac{d^2y}{dt^2} = -p^2(A \cos pt + B \sin pt)$
Step 4 (Substitute original y): Since $y = A \cos pt + B \sin pt$, we get:
$\frac{d^2y}{dt^2} = -p^2y \implies \frac{d^2y}{dt^2} + p^2y = 0$. (Proved)
Example 2
If $y = 2e^{3x} + 3e^{-2x}$, prove that $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y = 0$.
જો $y = 2e^{3x} + 3e^{-2x}$ હોય તો સાબિત કરો કે $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y = 0$.
Step 1 (First Derivative): $\frac{dy}{dx} = y_1 = 2(3e^{3x}) + 3(-2e^{-2x}) = 6e^{3x} - 6e^{-2x}$
Step 2 (Second Derivative): $\frac{d^2y}{dx^2} = y_2 = 6(3e^{3x}) - 6(-2e^{-2x}) = 18e^{3x} + 12e^{-2x}$
Step 3 (Evaluate LHS): Substitute $y_2$, $y_1$, and $y$ into LHS = $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 6y$.
LHS = $(18e^{3x} + 12e^{-2x}) - (6e^{3x} - 6e^{-2x}) - 6(2e^{3x} + 3e^{-2x})$
Step 4 (Group and Cancel):
For $e^{3x}$: $18 - 6 - 12 = 0$
For $e^{-2x}$: $12 - (-6) - 18 = 12 + 6 - 18 = 0$.
LHS = $0 =$ RHS. (Proved)
5. Applications of Differentiation
વિકલનના ઉપયોગો
Part A: Velocity & Acceleration (વેગ અને પ્રવેગ)
Example 1
The equation of motion is $s = t^3 - 6t^2 + 9t$. Find velocity and acceleration at $t=3$ sec. Find $t$ when acceleration is zero.
એક કણની ગતિનું સમીકરણ $s = t^3 - 6t^2 + 9t$ છે. $t=3$ સેકન્ડે તેનો વેગ અને પ્રવેગ મેળવો. જ્યારે પ્રવેગ શૂન્ય હોય ત્યારે $t$ શોધો.
Step 1 (Velocity equation): Velocity is $v = \frac{ds}{dt} = 3t^2 - 12t + 9$.
Step 2 (Velocity at t=3): Substitute $t=3$. $v = 3(3)^2 - 12(3) + 9 = 27 - 36 + 9 = 0$ units/sec.
Step 3 (Acceleration equation): Acceleration is $a = \frac{dv}{dt} = 6t - 12$.
Step 4 (Acceleration at t=3): Substitute $t=3$. $a = 6(3) - 12 = 18 - 12 = 6$ units/sec².
Step 5 (Time when a=0): Set $a = 0 \implies 6t - 12 = 0 \implies 6t = 12 \implies t = 2$ seconds.
Example 2
The equation of motion of a particle is $s = 2t^3 - 3t^2 - 12t + 7$. Find velocity when acceleration is zero.
કણની ગતિનું સમીકરણ $s = 2t^3 - 3t^2 - 12t + 7$ છે. જ્યારે પ્રવેગ શૂન્ય હોય ત્યારે વેગ શોધો.
Step 1 (Find v and a):
Velocity $v = \frac{ds}{dt} = 6t^2 - 6t - 12$.
Acceleration $a = \frac{dv}{dt} = 12t - 6$.
Step 2 (Set a = 0): To find when acceleration is zero, set $12t - 6 = 0$.
$\implies 12t = 6 \implies t = 0.5$ seconds.
Step 3 (Find Velocity): Substitute $t=0.5$ back into the velocity equation.
$v = 6(0.5)^2 - 6(0.5) - 12 = 6(0.25) - 3 - 12$
$v = 1.5 - 3 - 12 = -13.5$ units/sec.
Example 3
The distance $s$ at time $t$ is given by $s = t^3 - 5t^2 + 3t + 4$. Find the time when the velocity is zero.
$s = t^3 - 5t^2 + 3t + 4$ માટે જ્યારે વેગ શૂન્ય હોય તે સમય શોધો.
Step 1 (Find Velocity): Differentiate $s$ with respect to $t$.
$v = \frac{ds}{dt} = 3t^2 - 10t + 3$.
Step 2 (Set v = 0): Since we want the time when velocity is zero, set the equation to zero:
$3t^2 - 10t + 3 = 0$.
Step 3 (Solve Quadratic Equation): Factorize the equation by splitting the middle term.
$3t^2 - 9t - 1t + 3 = 0$
$3t(t - 3) - 1(t - 3) = 0$
$(3t - 1)(t - 3) = 0$
Step 4 (Final Answer): The velocity is zero at $t = 3$ seconds or $t = \frac{1}{3}$ seconds.
Part B: Maxima & Minima (મહત્તમ અને ન્યૂનતમ મૂલ્યો)
Example 4
Find maximum and minimum value of $f(x) = 2x^3 - 3x^2 - 12x + 5$.
વિધેય $f(x) = 2x^3 - 3x^2 - 12x + 5$ ની મહત્તમ અને ન્યૂનત્તમ કિંમત શોધો.
Step 1 (First Derivative): Find $f'(x) = 6x^2 - 6x - 12$.
Step 2 (Critical Points): Set $f'(x) = 0 \implies 6(x^2 - x - 2) = 0 \implies (x-2)(x+1) = 0$. Roots are $x=2$ and $x=-1$.
Step 3 (Second Derivative): Find $f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6$.
Step 4 (Test $x=2$): $f''(2) = 12(2) - 6 = 18$. Since $18 > 0$, function has a Minimum at $x=2$.
Min Value = $f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$.
Step 5 (Test $x=-1$): $f''(-1) = 12(-1) - 6 = -18$. Since $-18 < 0$, function has a Maximum at $x=-1$.
Max Value = $f(-1) = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$.
Example 5
Find the maximum and minimum values of $f(x) = x^3 - 3x + 11$.
$f(x) = x^3 - 3x + 11$ ના મહત્તમ અને ન્યૂનતમ મૂલ્યો શોધો.
Step 1 (First Derivative): $f'(x) = 3x^2 - 3$.
Step 2 (Critical Points): Set $f'(x) = 0 \implies 3(x^2 - 1) = 0 \implies x^2 = 1 \implies x = 1$ and $x = -1$.
Step 3 (Second Derivative): $f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x$.
Step 4 (Test $x=1$): $f''(1) = 6(1) = 6 > 0$, so it is a Minimum.
Min Value = $f(1) = (1)^3 - 3(1) + 11 = 1 - 3 + 11 = 9$.
Step 5 (Test $x=-1$): $f''(-1) = 6(-1) = -6 < 0$, so it is a Maximum.
Max Value = $f(-1) = (-1)^3 - 3(-1) + 11 = -1 + 3 + 11 = 13$.
Example 6
Find maximum and minimum values for the function $f(x) = x^3 - 6x^2 + 9x + 6$.
વિધેય $f(x) = x^3 - 6x^2 + 9x + 6$ ના મહત્તમ અને ન્યૂનતમ મૂલ્યો શોધો.
Step 1 (First Derivative): $f'(x) = 3x^2 - 12x + 9$.
Step 2 (Critical Points): Set $f'(x) = 0 \implies 3(x^2 - 4x + 3) = 0 \implies 3(x-3)(x-1) = 0$. Roots are $x=3$ and $x=1$.
Step 3 (Second Derivative): $f''(x) = 6x - 12$.
Step 4 (Test $x=1$): $f''(1) = 6(1) - 12 = -6 < 0$, so it is a Maximum.
Max Value = $f(1) = 1^3 - 6(1)^2 + 9(1) + 6 = 1 - 6 + 9 + 6 = 10$.
Step 5 (Test $x=3$): $f''(3) = 6(3) - 12 = 18 - 12 = 6 > 0$, so it is a Minimum.
Min Value = $f(3) = 3^3 - 6(3)^2 + 9(3) + 6 = 27 - 54 + 27 + 6 = 6$.
Example 7
Find the maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 10$.
$f(x) = 2x^3 - 15x^2 + 36x + 10$ ના મહત્તમ અને ન્યૂનતમ મૂલ્યો શોધો.
Step 1 (First Derivative): $f'(x) = 6x^2 - 30x + 36$.
Step 2 (Critical Points): Set $f'(x) = 0 \implies 6(x^2 - 5x + 6) = 0 \implies 6(x-2)(x-3) = 0$. Roots are $x=2$ and $x=3$.
Step 3 (Second Derivative): $f''(x) = 12x - 30$.
Step 4 (Test $x=2$): $f''(2) = 12(2) - 30 = 24 - 30 = -6 < 0$, so it is a Maximum.
Max Value = $f(2) = 2(8) - 15(4) + 36(2) + 10 = 16 - 60 + 72 + 10 = 38$.
Step 5 (Test $x=3$): $f''(3) = 12(3) - 30 = 36 - 30 = 6 > 0$, so it is a Minimum.
Min Value = $f(3) = 2(27) - 15(9) + 36(3) + 10 = 54 - 135 + 108 + 10 = 37$.