Definitions [12]
Let \[f\] be a real-valued function which is a composite of two functions \[u\] and \[v\]; i.e., \[f = v \circ u\]. Suppose \[t = u(x)\] and if both \[\frac{dt}{dx}\]and \[\frac{dv}{dt}\]exist, we have
If \(u = g(x)\) and \(y = f(u)\), then \(y = f(g(x))\) is called a composite function. Here, \(g(x)\) is the inner function and \(f(u)\) is the outer function.
If a function reverses the action of another function, it is called its inverse function. For example, if \[y = \sin^{-1} x\], then \[x = \sin y\], which means the inverse function converts a trigonometric value back into an angle.
If b > 0, \[b \neq 1\], and a > 0, then
This means a logarithm tells the exponent to which the base must be raised to obtain the number.
A function of the form \[y = b^x\], where b > 0 and \[b \neq 1\], is called an exponential function.
Implicit differentiation means differentiating both sides of an equation with respect to x, while remembering that y depends on x. Therefore, whenever a term containing y is differentiated, the factor \[\frac{dy}{dx}\] appears by the chain rule.
When x = f(t) and y = g(t), the relation between x and y is said to be in parametric form.
Let y = f(x). If the first derivative \[\frac{dy}{dx} = f'(x)\] is itself differentiable, then differentiating once again with respect to \[x\] gives the second order derivative.
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then
\[\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\]
If y = f(x) is a differentiable function of x such that the inverse function x = f − 1(y) exists, then x is a differentiable function of
y and
\[\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}},\frac{dy}{dx}\neq0\]
If x = f(t) and y = g(t) are differential functions of parameter ‘t’, then y is a differential function of x and
\[\begin{aligned}
\frac{dy}{dx} & =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \\
\\
\frac{dx}{dt} & \neq0
\end{aligned}\]
If y = f(x) is a differentiable function of x, then its derivative f′(x) is also a function of x.
If this derivative f′(x) is again differentiable, its derivative is called the second derivative of f(x).
\[f^{\prime\prime}(x)\quad\mathrm{or}\quad\frac{d^2y}{dx^2}\]
If the second derivative is differentiable, its derivative is called the third derivative, denoted by:
\[f^{\prime\prime\prime}(x)\quad\mathrm{or}\quad\frac{d^3y}{dx^3}\]
Continuing this process, the derivative obtained after differentiating f(x) n times is called the nth derivative of f(x), and is denoted by:
\[f^{(n)}(x)\quad\mathrm{or}\quad\frac{d^ny}{dx^n}\]
These derivatives beyond the first derivative are called higher-order derivatives.
Formulae [11]
| Function | Derivative |
|---|---|
| [f(x)]ⁿ | n[f(x)]ⁿ⁻¹ · f′(x) |
| \[\sqrt{\mathrm{f}(x)}\] | \[\frac{1}{2\sqrt{\mathrm{f}(x)}}\cdot\mathrm{f}^{\prime}(x)\] |
| \[\frac{1}{\mathrm{f}(x)}\] | \[-\frac{1}{\left[\mathrm{f}(x)\right]^{2}}\cdot\mathrm{f}^{\prime}(x)\] |
| sin(f(x)) | cos(f(x)) · f′(x) |
| cos(f(x)) | −sin(f(x)) · f′(x) |
| tan(f(x)) | sec²(f(x)) · f′(x) |
| cot(f(x)) | −cosec²(f(x)) · f′(x) |
| sec(f(x)) | sec(f(x)) tan(f(x)) · f′(x) |
| cosec(f(x)) | −cosec(f(x)) cot(f(x)) · f′(x) |
| \[\mathbf{a}^{\mathbf{f}(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| \[\mathrm{e}^{\mathrm{f}(x)}\] | \[\mathrm{e}^{\mathrm{f}(x)\cdot\mathrm{f}^{\prime}(x)}\] |
| log(f(x)) | \[\frac{1}{\mathrm{f}(x)}\cdot\mathrm{f}^{\prime}(x)\] |
| logₐ(f(x)) | \[\frac{1}{\mathrm{f}(x)\mathrm{loga}}\cdot\mathrm{f}^{\prime}(x)\] |
| Function | Derivative | Condition |
|---|---|---|
| sin⁻¹x | \[\frac{1}{\sqrt{1-x^{2}}}\] | |x| < 1 |
| sin⁻¹(f(x)) | \[\frac{1}{\sqrt{1-\{f\left(x\right)\}^{2}}}\frac{d}{dx}f\left(x\right)\] | |f(x)| < 1 |
| cos⁻¹x | \[-\frac{1}{\sqrt{1-x^{2}}}\] | x| < 1 |
| cos⁻¹(f(x)) | \[-\frac{1}{\sqrt{1-\left\{f\left(x\right)\right\}^{2}}}\frac{d}{dx}f(x)\] | |f(x)| < 1 |
| tan⁻¹x | \[\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| tan⁻¹(f(x)) | \[\frac{1}{1+\left\{f\left(x\right)\right\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| cot⁻¹x | \[-\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| cot⁻¹(f(x)) | \[-\frac{1}{1+\{f(x)\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| sec⁻¹x | \[\frac{1}{|x|\sqrt{x^{2}-1}}\] | |x| > 1 |
| sec⁻¹(f(x)) | \[\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| cosec⁻¹x | \[-\left(\frac{1}{|x|\sqrt{x^{2}-1}}\right)\] |
|x| > 1 |
| cosec⁻¹(f(x)) | \[-\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| Function | Derivative | Condition |
|---|---|---|
| sin⁻¹x | \[\frac{1}{\sqrt{1-x^{2}}}\] | |x| < 1 |
| sin⁻¹(f(x)) | \[\frac{1}{\sqrt{1-\{f\left(x\right)\}^{2}}}\frac{d}{dx}f\left(x\right)\] | |f(x)| < 1 |
| cos⁻¹x | \[-\frac{1}{\sqrt{1-x^{2}}}\] | x| < 1 |
| cos⁻¹(f(x)) | \[-\frac{1}{\sqrt{1-\left\{f\left(x\right)\right\}^{2}}}\frac{d}{dx}f(x)\] | |f(x)| < 1 |
| tan⁻¹x | \[\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| tan⁻¹(f(x)) | \[\frac{1}{1+\left\{f\left(x\right)\right\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| cot⁻¹x | \[-\left(\frac{1}{1+x^{2}}\right)\] | x ∈ R |
| cot⁻¹(f(x)) | \[-\frac{1}{1+\{f(x)\}^{2}}\frac{d}{dx}f(x)\] | f(x) ∈ R |
| sec⁻¹x | \[\frac{1}{|x|\sqrt{x^{2}-1}}\] | |x| > 1 |
| sec⁻¹(f(x)) | \[\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
| cosec⁻¹x | \[-\left(\frac{1}{|x|\sqrt{x^{2}-1}}\right)\] |
|x| > 1 |
| cosec⁻¹(f(x)) | \[-\frac{1}{|f(x)|\sqrt{\{f(x)\}^{2}-1}}\frac{d}{dx}f(x)\] | |f(x)| > 1 |
1. Sum Rule:
\[y=u\pm v\] then \[\frac{dy}{dx}=\frac{du}{dx}\pm\frac{dv}{dx}\]
2. Product Rule:
\[y=uv\] then \[\frac{dy}{dx}=u\frac{d\nu}{dx}+\nu\frac{du}{dx}\]
3. Quotient Rule:
\[y=\frac{u}{v}\] where v ≠ 0 then \[\frac{dy}{dx}=\frac{\nu\frac{du}{dx}-u\frac{d\nu}{dx}}{\nu^{2}}\]
4. Difference Rule:
y = u − v then \[\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}\]
5. Constant Multiple:
y = k. u then \[\frac{dy}{dx}=k.\frac{du}{dx}\], k constant.
| y = f(x) | \[\frac{dy}{dx}=f^{\prime}(x)\] |
|---|---|
| c (Constant) | 0 |
| \[X^{n}\] | \[nx^{n-1}\] |
| \[\frac{1}{x}\] | \[-\frac{1}{x^2}\] |
| \[\frac{1}{x^n}\] | \[-\frac{n}{x^{n+1}}\] |
| \[\sqrt{x}\] | \[\frac{1}{2\sqrt{x}}\] |
| sin x | cos x |
| cos x | -sin x |
| tan x | sec2 x |
| cot x | -cosec2 x |
| sec x | sec x.tan x |
| cosec x | -cosec x cot x |
| \[e^{X}\] | \[e^{X}\] |
| \[a^{X}\] | \[a^xloga\] |
| log x | \[\frac{1}{x}\] |
| \[\log_{a}x\] | \[\frac{1}{x\log a}\] |
| Type of Function | Derivative |
|---|---|
| \[a^{x}\] | \[a^x\log a\] |
| \[e^{x}\] | \[e^{x}\] |
| \[x^{x}\] | \[x^x(1+\log x)\] |
| \[x^{a}\](a constant) | \[ax^{a-1}\] |
| \[a^{f(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| Given | Formula / Result |
|---|---|
| x = f(t), ; y = g(t) | Parametric form |
| First derivative | \[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\] |
| Condition | \[\frac{dx}{dt}\neq0\] |
| Second derivative | \[\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)/\frac{dx}{dt}\] |
If: u = f(x),v = g(x)
Then: \[\frac{du}{dv}=\frac{du/dx}{dv/dx}\]
General implicit form: F(x,y) = 0
\[x^my^n=(x+y)^{m+n}\]
\[\frac{dy}{dx}=\frac{y}{x}\]
| Expression | Derivative |
|---|---|
| \[y^{n}\] | \[ny^{n-1}\frac{dy}{dx}\] |
| f (y) | \[f^{\prime}(y)\frac{dy}{dx}\] |
| sin y | \[\cos y\frac{dy}{dx}\] |
| cos y | \[-\sin y\frac{dy}{dx}\] |
| \[e^{y}\] | \[e^y\frac{dy}{dx}\] |
| log y | \[\frac{1}{y}\frac{dy}{dx}\] |
| y | dy/dx |
|---|---|
| \[[f(x)]^{n}\] | \[n\left[f(x)\right]^{n-1}\cdot f^{\prime}(x)\] |
| \[\sqrt{f(x)}\] | \[\frac{f^{\prime}(x)}{2\sqrt{f(x)}}\] |
| \[\frac{1}{[f(x)]^{n}}\] | \[-\frac{n\cdot f^{\prime}(x)}{[f(x)]^{n+1}}\] |
| sin [f(x)] | \[\cos[f(x)]\cdot f^{\prime}(x)\] |
| cos [f(x)] | \[-\sin\left[f(x)\right]\cdot f^{\prime}(x)\] |
| tan [f(x)] | \[\sec^2[f(x)]\cdot f^{\prime}(x)\] |
| sec [f(x)] | \[\sec\left[f(x)\right]\cdot\tan\left[f(x)\right]\cdot f^{\prime}(x)\] |
| cot [f(x)] | \[-\operatorname{cosec}^2[f(x)]\cdot f^{\prime}(x)\] |
| cosec [f(x)] | \[-\operatorname{cosec}\left[f(x)\right]\cdot\cot\left[f(x)\right]\cdot f^{\prime}(x)\] |
| \[a^{f(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| \[e^{f(x)}\] | \[e^{f(x)}\cdot f^{\prime}(x)\] |
| log [f(x)] | \[\frac{f^\prime(x)}{f(x)}\] |
| \[\log_{a}[f(x)]\] | \[\frac{f^{\prime}(x)}{f(x)\log a}\] |
| y | dy/dx | Conditions |
|---|---|---|
| \[\sin^{-1}x\] | \[\frac{1}{\sqrt{1-x^2}},|x|<1\] | −1 ≤ x ≤ 1 \[-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\] |
| \[\cos^{-1}x\] | \[-\frac{1}{\sqrt{1-x^{2}}},|x|<1\] | −1 ≤ x ≤ 1 0 ≤ y ≤ π |
| \[\tan^{-1}x\] | \[\frac{1}{1+x^2}\] | x ∈ R \[-\frac{\pi}{2}<y<\frac{\pi}{2}\] |
| \[\cot^{-1}x\] | \[-\frac{1}{1+x^2}\] | x ∈ R 0 < y < π |
| \[\sec^{-1}x\] | \[\frac{1}{x\sqrt{x^{2}-1}}\quad\mathrm{for}x>1\] | 0 ≤ y ≤ π |
| \[-\frac{1}{x\sqrt{x^2-1}}\mathrm{~for~}x<-1\] | \[y\neq\frac{\pi}{2}\] | |
| \[cosec^{-1}x\] | \[-\frac{1}{x\sqrt{x^{2}-1}}\mathrm{for}x>1\] | \[-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\] |
| \[{\frac{1}{x{\sqrt{x^{2}-1}}}}\quad{\mathrm{for}}x<-1\] | \[y\neq0\] |
Theorems and Laws [2]
If x = `e^(x/y)`, then prove that `dy/dx = (x - y)/(xlogx)`.
Given: x = `e^(x/y)`
Taking log on both the sides,
log x = `log e^(x/y)`
⇒ log x = `x/y log e`
⇒ log x = `x/y` ...[∵ log e = 1] ...(i)
Differentiating both sides w.r.t. x:
`d/dx log x = d/dx (x/y)`
⇒ `1/x = (y xx 1 - x xx dy/dx)/y^2`
⇒ `y^2 = xy - x^2 xx dy/dx`
⇒ `x^2 xx dy/dx = xy - y^2`
⇒ `dy/dx = (y(x - y))/x^2`
⇒ `dy/d = y/x xx ((x - y)/x)`
⇒ `dy/dx = 1/logx xx ((x - y)/x) ...[∵ log x = x/y "from equation (i)"]`
`dy/dx = (x - y)/(xlogx)`
Hence proved.
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.
Given, y = 5 cos x – 3 sin x
Differentiating both sides with respect to x,
`dy/dx = 5 d/dx cos x - 3 d/dx sin x`
= 5 (−sin x) − 3 cos x
= −5 sin x − 3 cos x
Differentiating both sides again with respect to x,
`(d^2 y)/dx = - 5 d/dx sin x - 3 d/dx cos x`
= −5 cos x − 3 (−sin x)
= 3 sin x − 5 cos x
Hence, `(d^2 y)/dx^2 + y` = 0
(3 sin x − 5 cos x) + (5 cos x − 3 sin x) = 0 ...(On substituting the value of y)
Key Points
-
A composite function has one function inside another function.
-
The chain rule formula is \[\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}\]
-
First differentiate the outer function, then multiply by the derivative of the inner function.
-
The derivative of an inverse function is usually found using implicit differentiation.
-
For \[\sin^{-1} x\] and \[\cos^{-1} x\], the denominator is \[\sqrt{1 - x^2}\].
-
For \[\tan^{-1} x\] and \[\cot^{-1} x\], the denominator is \[1 + x^2\].
-
For \[\sec^{-1} x\] and \[\csc^{-1} x\], the denominator involves \[|x|\sqrt{x^2 - 1}\].
-
Negative signs are especially important in \[\cos^{-1} x\], \[\cot^{-1} x\], and \[\csc^{-1} x\].
-
Domain restrictions must be checked before applying formulas.
If y = f(x) is a differentiable function of x such that the inverse function x = f⁻¹(y) exists, then x is a differentiable function of y and
\[\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}\], where \[\frac{\mathrm{d}y}{\mathrm{d}x}\neq0\].
-
Exponential function: \[y = b^x\], domain = all real numbers, range = positive real numbers.
-
Logarithmic function: \[y = \log_b x\], domain = positive real numbers, range = all real numbers.
-
Exponential and logarithmic functions are inverses of each other.
-
\[e^x\] and log x are especially important in calculus.
-
Main log laws: product, quotient, power, and change of base.
-
Standard derivatives: \[\frac{d}{dx}(e^x) = e^x\],
\[\frac{d}{dx}(\log x) = \frac{1}{x}\].
- If an equation contains both x and y and cannot be solved directly for y, it is called an implicit function.
- Implicit functions are generally written in the form:
f(x, y) = 0 - To differentiate an implicit function, differentiate both sides with respect to x, treating y as a function of x.
-
Parametric form means both x and y are written in terms of a third variable.
-
The third variable is called the parameter.
-
The main formula is:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\] -
This formula is based on the chain rule.
-
Always check that \[\frac{dx}{dt} \neq 0\].
-
The final answer may remain in terms of the parameter unless the question asks for conversion.
-
Second derivative means differentiating the function twice with respect to the same variable.
-
It is defined only when the first derivative is differentiable.
-
Common notations are \[\frac{d^2y}{dx^2}\], f''(x), y'', \[D^2y\], and \[y_2\].
-
Higher order derivatives can be defined similarly.
1. Elasticity of Demand
\[\eta=-\frac{P}{D}\cdot\frac{dD}{dP}\]
2. Marginal Revenue & Elasticity Relation
\[R_m=R_A\left(1-\frac{1}{\eta}\right)\]
3. Propensity to Consume & Save
MPC + MPS = 1
APC + APS = 1
Important Questions [10]
- Find dy/dx if y=cos^−1(√x)
- find dy/dx if y=tan^−1 (6x/(1−5x^2))
- If X Y = E X − Y , Show that D Y D X = Log X ( 1 + Log X ) 2
- The Total Cost Function of a Firm is C = X2 + 75x + 1600 for Output X. Find The Output for Which the Average Cost Ls Minimum. is Ca= Cm at this Output?
- If Y = Sin -1 ((8x)/(1 + 16x^2)), Find (Dy)/(Dx)
- Evaluate : Int (Sec^2 X)/(Tan^2 X + 4) Dx
- find dy/dx if x=e^(2t) , y=e^(√t)
- The Cost C of Producing X Articles is Given as C=X3-16x2+47x.
- If X7 . Y9 = (X + Y)16 Then Show that "Dy"/"Dx" = "Y"/"X"
- If X 3 Y 5 = ( X + Y ) 8 , Then Show that D Y D X = Y X
