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.
A function of the form \[y = b^x\], where b > 0 and \[b \neq 1\], is called an exponential function.
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.
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(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.
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then
\[\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\]
Formulae [10]
| 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 |
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.
-
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
