- Example 1 - Calculate the first derivative of afunction \(y = x^5 - 4x^3 + 2x -3.\) The derivative of the first element in \(y\) function (\(x^5\)) is solved using the first rule in the basic rules of function derivatives. $$x^5 = 5\cdot x^{5-1} = 5 \cdot x^4$$ The second element of y function \(-4x^3\) is solved using the first rule in the basic rules of function derivatives. $$4x^3 = 4\cdot 3 \cdot x^{3-1} = 12 \cdot x^2$$ The third element of the function \(y\) \(2\cdot x\)is solved using the first rule in the basic rules of function derivatives. $$ 2\cdot x = 2\cdot x^{1-1} = 2 \cdot x^0 = 2$$ The final element of the function \(y\) is a constant \(-3\). With application of the first rule of the basic rules of derivatives the derivation of the constant is equal to zero.
Solution: The first derivative of the function \(y\) with respect to \(x\) can be written as: $$y' = 5\cdot x^4 - 12\cdot x^2 + 2$$ - Example 2 - Calculate the first derivative of the following function \(y = \frac{1}{4} - \frac{1}{3}x + x^2 - 0.5x^4\). The first member of the function \(y\) is the consntant \(\frac{1}{4}\). The first deriviative of the constant is equal to zero. $$ \left(\frac{1}{4}\right)' = 0.$$ The second member of the function \(y\) is the \(-\frac{1}{3}x\) that consist of negative constant mutliplied by the variable \(x\). According to the rule number 4 of the basic rules of derivatives the first derivative of the constant multiplied by the variable is equal to the the constant multiplied by the first derivative of variable i.e. \((cy)' = xy'.\) With application of this rule to the second member of the function \(y\) $$\left(-\frac{1}{3}\cdot x\right)' = -\frac{1}{3}\cdot x^{1-1} = -\frac{1}{3}\cdot x^0 = -\frac{1}{3}.$$ The tird element of the function \(y\) is \(x^2\). With application of the first rule of the basic function of derivates the following solution for the third member is obtained $$ x^2 = 2 \cdot x^{2-1} = 2 \cdot x.$$ The first derivative of the fourth and final element of the function \(y\) can be written as $$ \left(-0.5 \cdot x^4\right)' = -0.5\cdot 4 \cdot x^{4-1} = -2\cdot x^3. $$ Solution - the first derivative of function \(y\) with respect to variable \(x\) can be written in the following form $$y' = 0-\frac{1}{3}\cdot x^{1-1} + 2\cdot 2\cdot x^{2-1} - 0.5 \cdot 4 \cdot x^{4-1} =-\frac{1}{3} + 2\cdot x - 2 \cdot x^2.$$
- Example 3 - Calculate the first derivative of the following function \(y = -\frac{5x^3}{a}\).
In this function the number 5 and letter \(a\) are treated as the constants and the only variable is \(x^3\). The first derivatiave of the \(x^3\) can be determined using the first rule of the basic function derivates. $$ (x^3)' = 3 \cdot x^{3-1} = 3 \cdot x^2$$ Solution - the first derivative of the function \(y\) can be written as: $$y' = -\frac{5\cdot 3 \cdot x^{3-1}}{a} = -\frac{15x^2}{a},$$ - Example 4 - Calculate the first derivative of the function: \(y = at^m + bt^{m+n}\). The first element of the function \(y\) is the constant \(a\) multiplied with variable \(t\) to the power of \(m\). The derivation of this component with repsect to \(t\) is done with application of the first rule of basic function derivatives \(x^n = n\cdot x^{n-1}\), and the first and fourht rule of the basic rules of derivatives i.e \((c)' = 0,\) and \((cy)' = cy'.\) The derivation of the first component of the function \(y\) with respect to \(t\) can be written in the following form $$ (a\cdot t^m)' = a\cdot m \cdot t^{m-1}.$$ The derivtion of the second component of the function \(y\) with respect to \(t\) is solved by using same derivation rules that were used for the first componenet. The result of application of those rules to the second element can be wrriten in the following form: $$ (b\cdot t^{m+n})' = b \cdot (m+n) \cdot t^{m+n-1}$$. The first derivation of the function \(y\) with respect to the variable \(t\) can be written as $$y' = amt^{m-1} + b(m+n)t^{m+n-1} = t^{m-1} \left(am + \frac{b(m+n)}{t}\right)$$
- Example 5 - Calculate the first derivative of the function: \(y=\frac{ax^6 + b}{\sqrt{a^2 + b^2}}\) with respect to variable \(x\)As seen in function \(y\) the only variable in this case is located at numerator \(x^6\) while letters \(a\) and \(b\) are constants. The derivation rules applied for deriving the numerator are
- Fourth rule of the basic rules of derivatives for constant \(a\) i.e \((c\cdot y)' = cy'\),
- First rule of basic function derivates for variable \(x^6\) i.e. \((x^6)' = 6\cdot x^5\)
- First rule of the basic rules of derivatives for constant \(b\) in the numerator
Solution - the first derivative of the function \(y\) with respect to the variable \(x\) can be written as: $$y' = \frac{6ax^5}{\sqrt{a^2+b^2}}.$$ - Example 6 - Calulate the first derivative of the function: \(y = \frac{\pi}{x} + \ln 2\) with resepct to \(x\).
In this case the variable x is located as denuminator in the frist member of the function \(y\) while the nominator of the first member and the second member are constants. The variable x can be written as $$ \frac{\pi}{x} = \pi\cdot x^{-1} $$ This variable can be derivated with application of the first rule for the derivates of basic functions $$\pi \cdot x^{-1} = -\pi \cdot x^ {-1-1} = -\pi \cdot x^{-2} = -\frac{\pi}{x^{2}}.$$
Solution - the first derivative of function \(y\) with respect to the variable \(x\) can be written as $$ y' = (\frac{\pi}{x} + \ln 2)'= -\frac{\pi}{x^2} + 0 = -\frac{\pi}{x^2}.$$ - Example 7 - Calculate the first derivative of the function:\(y = 3x^{\frac{2}{3}} - 2x^{\frac{5}{2}} + x^{-3}\) with respect to the vriable \(x\). The frist component can be derivated with respect to \(x\) as: $$ \left(3\cdot x^{\frac{2}{3}}\right)' = 3 \cdot \frac{2}{3} \cdot x^{\frac{2}{3}-1 } = 2 \cdot x^{\frac{2-3}{3}} = 2 \cdot x^{-\frac{1}{3}}.$$ The derivation of the second component of function \(y\) with respect to \(x\) can be written as: $$ \left(-2\cdot x^{\frac{5}{2}}\right)' = -2\cdot \frac{5}{2} \cdot x^{\frac{5}{2}-1} = -5 \cdot x^{\frac{5-2}{2}} = -5 \cdot x^{\frac{3}{2}}.$$ The derivation of the third component of function \(y\) with respect to \(x\) can be written as: $$\left(x^{-3}\right)' = -3 \cdot x^{-3-1} = -3 \cdot {-4}.$$ Solution
$$ y' =(3\cdot x^{\frac{2}{3}} - 2\cdot x^{\frac{5}{2}} + x^{-3})' = 2\cdot x^{-\frac{1}{3}} - 5\cdot x^{\frac{3}{2}} - 3\cdot x^{-4}$$ - Example 8 - Calculate the first derivative of the function \(y = x^2 \sqrt[3]{x^2} = x^\frac{8}{3}\) with respect to the variable \(x.\)
First the function \(y\) must be simplified. $$ y = x^2 \cdot x^{\frac{2}{3}} = x^{2+\frac{2}{3}} = x^{\frac{6 + 2}{3}} = x^{\frac{8}{3}}$$ Now that the functio is simplified the first basic function derivation rule is applied and the solution can be written as: $$ y' = \left(x^{\frac{8}{3}}\right)' = \frac{8}{3} x^{\frac{8-3}{3}} = \frac{8}{3} x^{\frac{5}{3}}$$ - Example 9 - Calculate the first derivative of the function\(y = \frac{a}{\sqrt[3]{x^2}} - \frac{b}{x\sqrt[3]{x}}\) with respect to the variable \(x\).
Before derivating the function with respect to the varialbe \(x\) first the function should be simplified. $$ y = a\cdot x^{-\frac{2}{3}} - b\cdot x^{-1} x^{-\frac{1}{3}} = a \cdot x^{-\frac{2}{3}} - b \cdot x^{\frac{-3-1}{3}} = a \cdot x^{-\frac{2}{3}} - b \cdot x^{-\frac{4}{3}}.$$ Now that the function is simplified the first derivation of the function y with respect to the variable x can be solve with application of the fourth rule of the basic rules of derivatives and the first rule of the basic function derivates. Solution can be written in the following form: $$ \begin{eqnarray}y' &=& \left(a \cdot x^{-\frac{2}{3}} - b \cdot x^{-\frac{4}{3}}\right)'\\\nonumber y'&=& -\frac{2}{3}\cdot a \cdot x^{-\frac{2}{3} - 1} + b \cdot \frac{4}{3} \cdot x^{-\frac{4}{3}-1} \\ \nonumber y' &=& -\frac{2}{3}\cdot a \cdot x^{\frac{-2-3}{3}} + b \cdot \frac{4}{3} \cdot x^{\frac{-4-3}{3}}\\ \nonumber y' &=& -\frac{2}{3}\cdot a \cdot x^{\frac{-5}{3}} + b \cdot \frac{4}{3} \cdot x^{\frac{-7}{3}}\end{eqnarray}$$
The Derivative of Algebraic Functions (with examples)
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