Definitions [6]
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.
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
\[ \boxed{\dfrac{df}{dx} = \dfrac{dv}{dt} \cdot \dfrac{dt}{dx}} \]
For example, if \[ f = w \circ v \circ u, \]
then the derivative is obtained by multiplying the successive derivatives.
\[ \boxed{\dfrac{df}{dx} = \dfrac{dw}{ds} \cdot \dfrac{ds}{dt} \cdot \dfrac{dt}{dx}} \]
where the intermediate variables represent the nested functions.
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 differentiation of a function is performed after taking logarithm on both sides, the process is called logarithmic differentiation.
This method is especially useful when:
- both the base and exponent contain variables;
- the function contains complicated products, quotients, or powers.
General Form:
\[ \boxed{\dfrac{dy}{dx} = y\left[\dfrac{v(x)}{u(x)}u'(x) + v'(x)\log[u(x)]\right]} \]
Since \[ y = [u(x)]^{v(x)}, \]
we may also write
\[ \boxed{\dfrac{dy}{dx} = [u(x)]^{v(x)}\left[\dfrac{v(x)}{u(x)}u'(x) + v'(x)\log[u(x)]\right]} \]
Implicit 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.
Explicit Function
If a relation between x and y can be easily solved for y and written as \[ y = f(x), \] then y is given as an explicit function of x.
When x and y are expressed separately as functions of the same third variable t, i.e. \[ x = f(t), \qquad y = g(t), \] the equations are called parametric equations, and t is called the parameter.
The parameter may also be denoted by \[\theta, u,\] etc.
Important Condition
The formula \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \] is directly applicable when
\[ \boxed{\frac{dx}{dt} \neq 0.} \]
If \[ \frac{dx}{dt} = 0, \] the point must be examined separately. It may correspond to a point where the tangent is vertical.
Formulae [3]
| y = f(x) | dy/dx = f′(x) |
|---|---|
| c (Constant) | 0 |
| xⁿ | n xⁿ⁻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 |
| sec x | sec x tan x |
| cosec x | −cosec x cot x |
| cot x | −cosec² x |
| eˣ | eˣ |
| aˣ | aˣ log a |
| log x | \[\frac{1}{x}\] |
| logₐ x | \[\frac{1}{x\log a}\] |
| 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) |
| \[\mathbf{a}^{\mathbf{f}(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| \[\mathrm{e}^{\mathrm{f}(x)}\] | \[ e^{f(x)} f'(x) \] |
| 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 |
Theorems and Laws [1]
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that `dy/dx = cos^2(a+y)/(sin a)`.
cos y = x cos (a + y)
∴ x = `(cos y)/(cos (a + y))`
On differentiating with respect to y,
`cos (a + y) d/dy cos y - cos y d/dy`
`therefore dx/dy = (cos (a + y))/(cos^2 (a + y))`
`= (- sin y cos (a + y) + cos y sin (a + y))/(cos^2 (a + y))`
`= (sin (a + y) cos y - cos (a + y) sin y)/(cos^2 (a + y))`
`= (sin (a + y - y))/(cos^2 (a + y))` ...[∵ sin (A − B) = sin A cos B − cos A sin B]
`= (sin a)/(cos^2 (a + y))`
`therefore dy/dx = (cos^2 (a + y))/(sin a)`
Key Points
Let f(x), g(x) and h(x) are three real valued functions are given, then
(i) Sum of two functions
\[[f(x)+g(x)]^{\prime}=f^{\prime}(x)+g^{\prime}(x)\]
(ii) Difference of two functions
\[[f(x)-g(x)]^{\prime}=f^{\prime}(x)-g^{\prime}(x)\]
(iii) Product of two or more functions
(a) \[[f(x)\cdot g(x)]^{\prime}=f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x)\]
(b) \[\frac{d}{dx}[f(x)\cdot g(x)\cdot h(x)]=f(x)\cdot g(x)\cdot h^{\prime}(x)\]\[+f\left(x\right)\cdot g^{\prime}\left(x\right)\cdot h(x)+f^{\prime}\left(x\right)\cdot g\left(x\right)\cdot h\left(x\right)\]
(iv) Quotient of two functions
\[\left[\frac{f(x)}{g(x)}\right]'\] = \[\frac{f^{\prime}(x)\cdot g(x)-g^{\prime}(x)\cdot f(x)}{\left[g(x)\right]^{2}}\]
provided g(x) ≠ 0
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A composite function has one function inside another function.
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The chain rule formula is \[\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}\]
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First differentiate the outer function, then multiply by the derivative of the inner function.
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 \[\frac{\mathrm{d}y}{\mathrm{d}x}\neq0\].
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The derivative of an inverse function is usually found using implicit differentiation.
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For \[\sin^{-1} x\] and \[\cos^{-1} x\], the denominator is \[\sqrt{1 - x^2}\].
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For \[\tan^{-1} x\] and \[\cot^{-1} x\], the denominator is \[1 + x^2\].
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For \[\sec^{-1} x\] and \[\csc^{-1} x\], the denominator involves \[|x|\sqrt{x^2 - 1}\].
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Negative signs are especially important in \[\cos^{-1} x\], \[\cot^{-1} x\], and \[\csc^{-1} x\].
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Domain restrictions must be checked before applying formulas.
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Use logarithmic differentiation when the function is a complex product, quotient, or variable exponent form.
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Write y = function first, then take \[\ln\] on both sides.
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Apply logarithmic rules before differentiating.
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Differentiate \[\ln y\] carefully: \[\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}\].
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Substitute the original value of y at the end.
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Ensure the expression inside logarithm remains positive.
- 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.
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Parametric form means both x and y are written in terms of a third variable.
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The third variable is called the parameter.
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The main formula is:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\] -
This formula is based on the chain rule.
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Always check that \[\frac{dx}{dt} \neq 0\].
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The final answer may remain in terms of the parameter unless the question asks for conversion.
- If y = f(x), then \[\frac{dy}{dx}\] = f′(x) is called the first-order derivative.
- The derivative of the first derivative is called the second-order derivative:
\[\frac{d^2y}{dx^2}\] = f″(x) - Higher order derivatives are written as:
fⁿ(x) or \[\frac{d^ny}{dx^n}\]
Concepts [10]
- Introduction & Derivatives of Some Standard Functions
- Algebra of Differentiation
- Derivatives of Composite Functions
- Geometrical Meaning of Derivative
- Derivative of Inverse Trigonometric Function
- Logarithmic Differentiation
- Differentiation of Implicit Functions
- Derivatives of Functions in Parametric Forms
- Higher Order Derivatives
- Successive Differentiation
