Answer
385.5k+ views
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, \[\dfrac{df}{dx}=\displaystyle \lim_{u \to x}\dfrac{{{u}^{n}}-{{x}^{n}}}{u-x}=n{{x}^{n-1}}\].
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, \[\dfrac{df}{dx}=\displaystyle \lim_{u \to x}\dfrac{{{u}^{n}}-{{x}^{n}}}{u-x}=n{{x}^{n-1}}\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)