Answer
Verified
412.2k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE