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How do you find the derivative of $\left( \dfrac{1+3x}{3x} \right)\left( 3-x \right)$?

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Last updated date: 12th Jul 2024
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Answer
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Hint: We explain the concept of derivation of a dependent variable with respect to an independent variable. We first find the formula for the derivation for ${{n}^{th}}$ power of a variable x where $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. We place the value for $n=-1,1$. We get the solution for the derivative of $f\left( x \right)=\dfrac{1}{x}$. We also explain the theorem with the help of the first order derivative.

Complete step-by-step solution:
Differentiation, the fundamental operations in calculus deals with the rate at which the dependent variable changes with respect to the independent variable. The measurement quantity of its rate of change is known as derivative or differential coefficients. We find the increment of those variables for small changes. We mathematically express it as $\dfrac{dy}{dx}$ where $y=f\left( x \right)$.
We need to find the derivative of $\left( \dfrac{1+3x}{3x} \right)\left( 3-x \right)$. We simplify the expression.
\[\left( \dfrac{1+3x}{3x} \right)\left( 3-x \right)=\dfrac{3-x}{3x}+\left( 3-x \right)=\dfrac{1}{x}-\dfrac{1}{3}+3-x\]. We need to find differentiation of constants which gives 0.
Let’s assume $y=f\left( x \right)=\dfrac{1}{x}-\dfrac{1}{3}+3-x$.
So, $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{1}{x}-\dfrac{1}{3}+3-x \right)=\dfrac{d}{dx}\left( \dfrac{1}{x} \right)+\dfrac{d}{dx}\left( -\dfrac{1}{3}+3 \right)+\dfrac{d}{dx}\left( -x \right)$.
Differentiating we get $\dfrac{dy}{dx}=\dfrac{-1}{{{x}^{2}}}-1$.
Therefore, the derivative of $\left( \dfrac{1+3x}{3x} \right)\left( 3-x \right)$ is $\dfrac{-1}{{{x}^{2}}}-1$.

Note: If the ratio of $\dfrac{\Delta y}{\Delta x}$ tends to a definite finite limit when \[\Delta x \to 0\], then the limiting value obtained by this can also be found by first order derivatives. We can also apply first order derivative theorem to get the differentiated value.
We know that $\dfrac{dy}{dx}=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Here $f\left( x \right)={{x}^{n}}$. Also, $f\left( x+h \right)={{\left( x+h \right)}^{n}}$. We assume $x+h=u$ which gives $f\left( u \right)={{\left( u \right)}^{n}}$ and $h=u-x$. As $h\to 0$ we get $u \to x$.
So, $\dfrac{df}{dx}=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\displaystyle \lim_{u \to x}\dfrac{{{u}^{n}}-{{x}^{n}}}{u-x}$.
We know the limit value $\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$.
Therefore, \[\dfrac{df}{dx}=\displaystyle \lim_{u \to x}\dfrac{{{u}^{n}}-{{x}^{n}}}{u-x}=n{{x}^{n-1}}\].