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
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.
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.
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) |
| \[\mathbf{a}^{\mathbf{f}(x)}\] | \[a^{f(x)}\log a\cdot f^{\prime}(x)\] |
| \[\mathrm{e}^{\mathrm{f}(x)}\] | \[\mathrm{e}^{\mathrm{f}(x)\cdot\mathrm{f}^{\prime}(x)}\] |
| 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 |
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.
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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 Function
- Logarithmic Differentiation
- Derivative of Implicit Functions
- Derivatives of Functions in Parametric Forms
- Higher Order Derivatives
- Successive Differentiation
