
If \[y=3{{x}^{5}}+4{{x}^{4}}+2x+3\], then
\[1)\] \[{{y}_{4}}=0\]
\[2)\]\[{{y}_{5}}=0\]
\[3)\]\[{{y}_{6}}=0\]
\[4)\]none of these
Answer
503.1k+ views
Hint : In this type of question \[{{y}_{i}}\] is the \[{{i}^{th}}\]order of derivative. So we just need to derivate of all the order given in the options of the question and check whether the option is correct or not.
To differentiate further we will use different ways like Chain Rule, Product Rule and various others.
Complete step-by-step solution:
When we want to find the different derivatives of any function then we differentiate that particular function with respect to the \[x\].Differentiation can be defined as the process of finding the derivatives, or rate of change of a function.
There are four partial types of rules-
Product Rule
Quotient Rule
Power Rule
Chain Rule
The chain rule provides a way to compute derivatives of a composite function. We use the chain rule when differentiating a ‘function of a function’ like\[f(g(x))\].
We use the product rule when the two differentiating functions are multiplied together, like \[f(x)g(x)\] in general. The product rule says that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.
As we have given in the question: \[y=3{{x}^{5}}+4{{x}^{4}}+2x+3\]
We need to find the derivative when the given \[y=3{{x}^{5}}+4{{x}^{4}}+2x+3\] becomes zero
Firstly we will find the first derivative i.e. \[{{y}_{1}}\]
So, for \[{{y}_{1}}\]we will differentiate with respect to \[x\]
\[{{y}_{1}}=\dfrac{d}{dx}(3{{x}^{5}}+4{{x}^{4}}+2x+3)\]
\[{{y}_{1}}=15{{x}^{4}}+16{{x}^{3}}+2\]
Now we will find the second derivative i.e. \[{{y}_{2}}\]
So,\[{{y}_{2}}=\dfrac{d}{dx}(15{{x}^{4}}+16{{x}^{3}}+2)\]
\[{{y}_{2}}=60{{x}^{3}}+48{{x}^{2}}\]
Now the third derivative i.e. \[{{y}_{3}}\] is given as:
\[{{y}_{3}}=\dfrac{d}{dx}(60{{x}^{3}}+48{{x}^{2}})\]
\[{{y}_{3}}=180{{x}^{2}}+96x\]
The fourth derivative \[{{y}_{4}}\]is given as:
\[\begin{align}
& {{y}_{4}}=\dfrac{d}{dx}(180{{x}^{2}}+96x) \\
& {{y}_{4}}=360x \\
\end{align}\]
As, it’s still not our required answer so we will find further derivatives-
Fifth derivative \[{{y}_{5}}\]is given as:
\[\begin{align}
& {{y}_{5}}=\dfrac{d}{dx}(360x) \\
& {{y}_{5}}=360 \\
\end{align}\]
Now we need to find the sixth derivative of the given function:
\[\begin{align}
& {{y}_{6}}=\dfrac{d}{dx}(360) \\
& {{y}_{6}}=0 \\
\end{align}\]
As we can observe that the sixth derivative is \[0\].
So, from all the above options, option \[(3)\] is correct.
Hence the final answer we got is option\[(3)\].
Note:With the help of differentiation we can calculate the highest and lowest point of the curve in a graph or we can also know its turning point. Newton’s method that is used for solving equations in computers, also uses the method of differentiation in it.
To differentiate further we will use different ways like Chain Rule, Product Rule and various others.
Complete step-by-step solution:
When we want to find the different derivatives of any function then we differentiate that particular function with respect to the \[x\].Differentiation can be defined as the process of finding the derivatives, or rate of change of a function.
There are four partial types of rules-
Product Rule
Quotient Rule
Power Rule
Chain Rule
The chain rule provides a way to compute derivatives of a composite function. We use the chain rule when differentiating a ‘function of a function’ like\[f(g(x))\].
We use the product rule when the two differentiating functions are multiplied together, like \[f(x)g(x)\] in general. The product rule says that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.
As we have given in the question: \[y=3{{x}^{5}}+4{{x}^{4}}+2x+3\]
We need to find the derivative when the given \[y=3{{x}^{5}}+4{{x}^{4}}+2x+3\] becomes zero
Firstly we will find the first derivative i.e. \[{{y}_{1}}\]
So, for \[{{y}_{1}}\]we will differentiate with respect to \[x\]
\[{{y}_{1}}=\dfrac{d}{dx}(3{{x}^{5}}+4{{x}^{4}}+2x+3)\]
\[{{y}_{1}}=15{{x}^{4}}+16{{x}^{3}}+2\]
Now we will find the second derivative i.e. \[{{y}_{2}}\]
So,\[{{y}_{2}}=\dfrac{d}{dx}(15{{x}^{4}}+16{{x}^{3}}+2)\]
\[{{y}_{2}}=60{{x}^{3}}+48{{x}^{2}}\]
Now the third derivative i.e. \[{{y}_{3}}\] is given as:
\[{{y}_{3}}=\dfrac{d}{dx}(60{{x}^{3}}+48{{x}^{2}})\]
\[{{y}_{3}}=180{{x}^{2}}+96x\]
The fourth derivative \[{{y}_{4}}\]is given as:
\[\begin{align}
& {{y}_{4}}=\dfrac{d}{dx}(180{{x}^{2}}+96x) \\
& {{y}_{4}}=360x \\
\end{align}\]
As, it’s still not our required answer so we will find further derivatives-
Fifth derivative \[{{y}_{5}}\]is given as:
\[\begin{align}
& {{y}_{5}}=\dfrac{d}{dx}(360x) \\
& {{y}_{5}}=360 \\
\end{align}\]
Now we need to find the sixth derivative of the given function:
\[\begin{align}
& {{y}_{6}}=\dfrac{d}{dx}(360) \\
& {{y}_{6}}=0 \\
\end{align}\]
As we can observe that the sixth derivative is \[0\].
So, from all the above options, option \[(3)\] is correct.
Hence the final answer we got is option\[(3)\].
Note:With the help of differentiation we can calculate the highest and lowest point of the curve in a graph or we can also know its turning point. Newton’s method that is used for solving equations in computers, also uses the method of differentiation in it.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

