
If \[y = {5^x}{x^5}\], then \[\dfrac{{dy}}{{dx}}\] is equal to
A) \[{5^x}\left( {{x^5}\log 5 - 5{x^4}} \right)\]
B) \[{x^5}\log 5 - 5{x^4}\]
C) \[{x^5}\log 5 + 5{x^4}\]
D) \[{5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)\]
Answer
412.2k+ views
Hint: Here, the given question. We have to find the derivative or differentiated term of the function. For this, first consider the given, then differentiate \[v\] with respect to \[x\] by using a standard differentiation formula and use product rule for differentiation then on further simplification we get the required differentiation value.
Complete step by step answer:
Differentiation can be defined as a derivative of a function with respect to an independent variable
(or)
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Let \[y = f\left( x \right)\] be a function of. Then, the rate of change of “y” per unit change in “x” is given by \[\dfrac{{dy}}{{dx}}\].
The product rule is used to differentiate many functions where one function is multiplied by another. The formal definition of the rule is: \[\dfrac{d}{{dx}}\left( {uv} \right) = u \cdot \dfrac{{dv}}{{dx}} + v \cdot \dfrac{{du}}{{dx}}\]
Consider the given function
\[ \Rightarrow \,\,\,\,y = {5^x}{x^5}\]---------- (1)
Here, \[y\] is a dependent variable and \[x\] is an independent variable.
Now we have to differentiate this function with respect to \[x\]
\[ \Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{5^x}{x^5}} \right)\]
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{5^x}{x^5}} \right)\]
Apply a product rule of differentiation, then
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x} \cdot \dfrac{d}{{dx}}\left( {{x^5}} \right) + \dfrac{d}{{dx}}\left( {{5^x}} \right) \cdot {x^5}\] -----(2)
Using the standard differentiated formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and \[\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\log a\], then equation (2) becomes
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x} \cdot \left( {5{x^4}} \right) + \left( {{5^x}\log 5} \right) \cdot {x^5}\]
Take ‘\[{5^x}\]’ as common, then we have
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x}\left( {5{x^4} + \log 5 \cdot {x^5}} \right)\]
Or
\[\therefore \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)\]
Hence, the required differentiated value \[\dfrac{{dy}}{{dx}} = {5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)\]. Therefore, option (D) is the correct answer.
Note:
When differentiating the function or term the student must recognize the dependent and independent variable then differentiate the dependent variable with respect to the independent variable. Should know the product and quotient rule of differentiation and remember the standard differentiation formulas.
Complete step by step answer:
Differentiation can be defined as a derivative of a function with respect to an independent variable
(or)
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Let \[y = f\left( x \right)\] be a function of. Then, the rate of change of “y” per unit change in “x” is given by \[\dfrac{{dy}}{{dx}}\].
The product rule is used to differentiate many functions where one function is multiplied by another. The formal definition of the rule is: \[\dfrac{d}{{dx}}\left( {uv} \right) = u \cdot \dfrac{{dv}}{{dx}} + v \cdot \dfrac{{du}}{{dx}}\]
Consider the given function
\[ \Rightarrow \,\,\,\,y = {5^x}{x^5}\]---------- (1)
Here, \[y\] is a dependent variable and \[x\] is an independent variable.
Now we have to differentiate this function with respect to \[x\]
\[ \Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{5^x}{x^5}} \right)\]
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{5^x}{x^5}} \right)\]
Apply a product rule of differentiation, then
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x} \cdot \dfrac{d}{{dx}}\left( {{x^5}} \right) + \dfrac{d}{{dx}}\left( {{5^x}} \right) \cdot {x^5}\] -----(2)
Using the standard differentiated formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and \[\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\log a\], then equation (2) becomes
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x} \cdot \left( {5{x^4}} \right) + \left( {{5^x}\log 5} \right) \cdot {x^5}\]
Take ‘\[{5^x}\]’ as common, then we have
\[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x}\left( {5{x^4} + \log 5 \cdot {x^5}} \right)\]
Or
\[\therefore \,\,\,\,\dfrac{{dy}}{{dx}} = {5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)\]
Hence, the required differentiated value \[\dfrac{{dy}}{{dx}} = {5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)\]. Therefore, option (D) is the correct answer.
Note:
When differentiating the function or term the student must recognize the dependent and independent variable then differentiate the dependent variable with respect to the independent variable. Should know the product and quotient rule of differentiation and remember the standard differentiation formulas.
Recently Updated Pages
Master Class 12 Social Science: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Class 10 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
The gas that burns in oxygen with a green flame is class 12 chemistry CBSE

The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE
