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Derivative of a Constant

Topic: Differentiation BasicsChapter: Chapter 8
ddx(c)=0\frac{d}{dx}(c) = 0
This rule states that the derivative of any constant 'c' with respect to 'x' is always zero. Constants do not change, so their rate of change is zero.

Power Rule

Topic: Differentiation BasicsChapter: Chapter 8
ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}
The power rule is used to differentiate functions where 'x' is raised to a power 'n'. You bring the power down as a coefficient and reduce the power by one. This is one of the most frequently used rules.

Constant Multiple Rule

Topic: Differentiation BasicsChapter: Chapter 8
ddx(cf(x))=cddx(f(x))\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))
If a function is multiplied by a constant 'c', you can take the constant out and differentiate the function 'f(x)' separately. The constant remains as a multiplier.

Sum and Difference Rule

Topic: Differentiation BasicsChapter: Chapter 8
ddx(f(x)±g(x))=ddx(f(x))±ddx(g(x))\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))
This rule allows you to differentiate functions that are sums or differences of other functions. You can differentiate each term separately and then add or subtract the results.

Derivative of e^x

Topic: Differentiation BasicsChapter: Chapter 8
ddx(ex)=ex\frac{d}{dx}(e^x) = e^x
The derivative of the exponential function e^x with respect to x is itself, e^x. This function is unique in calculus for this property.

Derivative of log_e x (Natural Logarithm)

Topic: Differentiation BasicsChapter: Chapter 8
ddx(log⁡ex)=1x\frac{d}{dx}(\log_e x) = \frac{1}{x}
The derivative of the natural logarithm of x (also written as ln x) is 1/x. This is fundamental for problems involving logarithmic growth or decay.

Product Rule

Topic: Differentiation BasicsChapter: Chapter 8
ddx(u⋅v)=udvdx+vdudx\frac{d}{dx}(u \cdot v) = u\frac{dv}{dx} + v\frac{du}{dx}
Used when differentiating a product of two functions, 'u' and 'v'. You differentiate the first function while keeping the second constant, then differentiate the second while keeping the first constant, and add the results.

Quotient Rule

Topic: Differentiation BasicsChapter: Chapter 8
ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}
Applied when differentiating a function that is a ratio of two other functions, 'u' and 'v'. It involves differentiating the numerator and denominator in a specific order and dividing by the square of the denominator.

Chain Rule

Topic: Differentiation BasicsChapter: Chapter 8
dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
The chain rule is used for differentiating composite functions (a function within a function). If y is a function of u, and u is a function of x, then the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x.

Second Order Derivative

Topic: Differentiation BasicsChapter: Chapter 8
d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)
The second order derivative is the derivative of the first derivative. It is used to determine the concavity of a function and is crucial for identifying maxima and minima in optimization problems.

Sum/Difference Rule

Topic: Differentiation BasicsChapter: Chapter 8
ddx(f(x)±g(x))=ddx(f(x))±ddx(g(x))\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))
The derivative of a sum or difference of two functions is the sum or difference of their individual derivatives.

Chain Rule (Basic Form)

Topic: Differentiation BasicsChapter: Chapter 8
dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
Used to differentiate composite functions, where 'y' is a function of 'u', and 'u' is a function of 'x'. It states that the derivative of 'y' with respect to 'x' is the derivative of 'y' with respect to 'u' multiplied by the derivative of 'u' with respect to 'x'.

Future Value with Continuous Compounding

Topic: Time Value of MoneyChapter: Chapter 8
FV=PertFV = P e^{rt}
This formula calculates the future value (FV) of a principal amount (P) when interest is compounded continuously at an annual interest rate (r) for a time period (t). The constant 'e' is Euler's number, approximately 2.71828.

Present Value with Continuous Compounding

Topic: Time Value of MoneyChapter: Chapter 8
PV=FVe−rtPV = FV e^{-rt}
This formula determines the present value (PV) of a future value (FV) when interest is compounded continuously at an annual rate (r) over a time period (t). It is the inverse of the future value formula for continuous compounding.

Present Value of a Continuous Annuity

Topic: Time Value of MoneyChapter: Chapter 8
PV=∫0nCe−rtdt=Cr(1−e−rn)PV = \int_{0}^{n} C e^{-rt} dt = \frac{C}{r} (1 - e^{-rn})
This formula calculates the present value of a continuous stream of constant cash flows (C per unit of time) received over 'n' periods, discounted at a continuous interest rate 'r'. It involves integrating the present value of each infinitesimal cash flow.

Future Value of a Continuous Annuity

Topic: Time Value of MoneyChapter: Chapter 8
FV=∫0nCer(n−t)dt=Cr(ern−1)FV = \int_{0}^{n} C e^{r(n-t)} dt = \frac{C}{r} (e^{rn} - 1)
This formula determines the future value of a continuous stream of constant cash flows (C per unit of time) made over 'n' periods, compounded continuously at a rate 'r'. It involves integrating the future value of each infinitesimal cash flow.

Present Value of a Continuous Perpetuity

Topic: Time Value of MoneyChapter: Chapter 8
PV=∫0∞Ce−rtdt=CrPV = \int_{0}^{\infty} C e^{-rt} dt = \frac{C}{r}
This formula calculates the present value of an infinite stream of continuous cash flows (C per unit of time) discounted at a continuous rate 'r'. It is a special case of the continuous annuity formula where 'n' approaches infinity.

Definition of Derivative

Topic: Logical ReasoningChapter: Chapter 8
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
This formula defines the derivative of a function f(x) as the instantaneous rate of change of f(x) with respect to x. It represents the slope of the tangent line to the function's graph at a given point.

Power Rule for Differentiation

Topic: Logical ReasoningChapter: Chapter 8
ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}
Used to differentiate functions where the variable is raised to a constant power. 'n' is any real number.

Constant Rule for Differentiation

Topic: Logical ReasoningChapter: Chapter 8
ddx(c)=0\frac{d}{dx}(c) = 0
The derivative of any constant 'c' is always zero, as constants do not change.

Sum/Difference Rule for Differentiation

Topic: Logical ReasoningChapter: Chapter 8
ddx(f(x)±g(x))=ddx(f(x))±ddx(g(x))\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))
The derivative of a sum or difference of functions is the sum or difference of their individual derivatives.

Product Rule for Differentiation

Topic: Logical ReasoningChapter: Chapter 8
ddx(u⋅v)=udvdx+vdudx\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}
Used to differentiate a product of two functions, 'u' and 'v', both of which are functions of 'x'.

Quotient Rule for Differentiation

Topic: Logical ReasoningChapter: Chapter 8
ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
Used to differentiate a function that is a ratio (quotient) of two other functions, 'u' and 'v', where 'v' is not zero.

Chain Rule for Differentiation

Topic: Logical ReasoningChapter: Chapter 8
dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
Used to differentiate composite functions. If 'y' is a function of 'u', and 'u' is a function of 'x', then the chain rule helps find the derivative of 'y' with respect to 'x'.

Derivative of Exponential Function

Topic: Logical ReasoningChapter: Chapter 8
ddx(eax)=aeax\frac{d}{dx}(e^{ax}) = ae^{ax}
The derivative of an exponential function with base 'e' (Euler's number). For a=1, it simplifies to \frac{d}{dx}(e^x) = e^x.

Derivative of Logarithmic Function

Topic: Logical ReasoningChapter: Chapter 8
ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}
The derivative of the natural logarithm function, \ln x (logarithm to base 'e').

Marginal Cost/Revenue/Profit

Topic: Logical ReasoningChapter: Chapter 8
MC=dCdQ,MR=dRdQ,MP=dPdQMC = \frac{dC}{dQ}, \quad MR = \frac{dR}{dQ}, \quad MP = \frac{dP}{dQ}
These formulas define marginal cost (MC), marginal revenue (MR), and marginal profit (MP) as the first derivatives of total cost (C), total revenue (R), and total profit (P) functions with respect to quantity (Q). They represent the change in cost/revenue/profit from producing/selling one additional unit.

Conditions for Maxima/Minima (Optimization)

Topic: Logical ReasoningChapter: Chapter 8
For critical points: f′(x)=0For Maxima: f′′(x)<0For Minima: f′′(x)>0\text{For critical points: } f'(x) = 0 \\ \text{For Maxima: } f''(x) < 0 \\ \text{For Minima: } f''(x) > 0
To find the maximum or minimum values of a function, first find the critical points by setting the first derivative to zero. Then, use the second derivative test: a negative second derivative indicates a local maximum, and a positive second derivative indicates a local minimum.

Power Rule for Integration

Topic: Logical ReasoningChapter: Chapter 8
∫xndx=xn+1n+1+C(neq−1)\int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n eq -1)
Used to integrate functions where the variable is raised to a constant power. 'C' is the constant of integration, representing any fixed cost or initial value.

Integral of 1/x

Topic: Logical ReasoningChapter: Chapter 8
∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C
The integral of 1/x, which is a special case not covered by the power rule for integration.

Integral of Exponential Function

Topic: Logical ReasoningChapter: Chapter 8
∫eaxdx=1aeax+C\int e^{ax} dx = \frac{1}{a}e^{ax} + C
The integral of an exponential function with base 'e'. For a=1, it simplifies to \int e^x dx = e^x + C.

Total Cost/Revenue/Profit from Marginal

Topic: Logical ReasoningChapter: Chapter 8
C(Q)=∫MC(Q)dQ+KC(Q) = \int MC(Q) dQ + K
To find the total cost (C), total revenue (R), or total profit (P) function from their respective marginal functions (MC, MR, MP), one must integrate the marginal function. 'K' represents the fixed cost or initial value, which is the constant of integration.

Definite Integral (Fundamental Theorem of Calculus)

Topic: Logical ReasoningChapter: Chapter 8
∫abf(x)dx=F(b)−F(a)\int_a^b f(x) dx = F(b) - F(a)
The definite integral calculates the net change of a function F(x) over an interval [a, b], where F'(x) = f(x). It can represent the area under the curve of f(x) from 'a' to 'b'.
Chapter 8 Formulas — CA Foundation | Kartavvya