
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
509.4k+ 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
Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

The coating formed on the metals such as iron silver class 12 chemistry CBSE

Metals are refined by using different methods Which class 12 chemistry CBSE

What do you understand by denaturation of proteins class 12 chemistry CBSE

Assertion Nitrobenzene is used as a solvent in FriedelCrafts class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

