Question

Find the derivative of the function \[y={{x}^{2x+1}}\]?

Answer 1

Hint: This question is from the topic of calculus. In solving this question, we will first put log base ‘e’ (that is \[\ln \]) to the both sides of the equation which is given in the question. After that, we will use the formula of logarithms that is \[\ln {{x}^{a}}=a\ln x\]. After that, we will differentiate both sides of the equation. After that, we will use the formulas of differentiation and solve the further question. After that, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find the derivative of \[y={{x}^{2x+1}}\].
So, we have to differentiate the equation
\[y={{x}^{2x+1}}\]
Now, taking \[\ln \] (that is log base e or \[{{\log }_{e}}\]) to the both side of the equation, we can write
\[\Rightarrow \ln y=\ln {{x}^{2x+1}}\]
Now, using the formula of logarithms that is \[\ln {{x}^{a}}=a\ln x\], we can write the above equation as
\[\Rightarrow \ln y=\left( 2x+1 \right)\ln x\]
The above equation can also be written as
\[\Rightarrow \left( \ln y \right)=2x\ln x+\ln x\]
Now, let us differentiate both sides of the equation, we can write
\[\Rightarrow d\left( \ln y \right)=d\left( 2x\ln x+\ln x \right)\]
The above equation can also be written as
\[\Rightarrow d\left( \ln y \right)=d\left( 2x\ln x \right)+d\left( \ln x \right)\]
Now, using the formula of differentiation that is \[d\left( \ln x \right)=\dfrac{1}{x}dx\], we can write the above differentiation as
\[\Rightarrow \dfrac{1}{y}dy=d\left( 2x\ln x \right)+\dfrac{1}{x}dx\]
Now, using the formula of product rule of differentiation that is \[d\left( uv \right)=v\centerdot d\left( u \right)+u\centerdot d\left( v \right)\], we can write the above differentiation as
\[\Rightarrow \dfrac{1}{y}dy=\ln x\centerdot d\left( 2x \right)+2x\centerdot d\left( \ln x \right)+\dfrac{1}{x}dx\]
Now, using the formula of differentiation that is \[d\left( n\cdot x \right)=n\cdot dx\], where n is any constant, we can write
\[\Rightarrow \dfrac{1}{y}dy=\ln x\centerdot 2dx+2x\centerdot d\left( \ln x \right)+\dfrac{1}{x}dx\]
Using the formula of differentiation that is \[d\left( \ln x \right)=\dfrac{1}{x}dx\], we can write
\[\Rightarrow \dfrac{1}{y}dy=\ln x\centerdot 2dx+2x\centerdot \dfrac{1}{x}dx+\dfrac{1}{x}dx\]
Now, taking ‘dx’ as common in the right side of the equation, we can write
\[\Rightarrow \dfrac{1}{y}dy=\left( 2\ln x+\dfrac{2x}{x}+\dfrac{1}{x} \right)dx\]
Now, dividing both side of the equation by ‘dx’ and multiplying both side of the equation by ‘y’, we can write
\[\Rightarrow \dfrac{dy}{dx}=\left( 2\ln x+\dfrac{2x}{x}+\dfrac{1}{x} \right)y\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dx}=\left( \dfrac{2x\ln x}{x}+\dfrac{2x}{x}+\dfrac{1}{x} \right)y=\left( \dfrac{2x\ln x+2x+1}{x} \right)y\]
Using the formula \[\ln {{x}^{a}}=a\ln x\], we can write the above equation as
\[\Rightarrow \dfrac{dy}{dx}=\left( \dfrac{\ln {{x}^{2x}}+2x+1}{x} \right)y\]
\[\Rightarrow \dfrac{dy}{dx}=\left( \ln {{x}^{2x}}+2x+1 \right)\dfrac{y}{x}\]
Now, putting the value of y in the above equation, we get
\[\Rightarrow \dfrac{dy}{dx}=\left( \ln {{x}^{2x}}+2x+1 \right)\dfrac{{{x}^{2x+1}}}{x}\]
Now, using the formula \[\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}\], we can write
\[\Rightarrow \dfrac{dy}{dx}=\left( \ln {{x}^{2x}}+2x+1 \right)\dfrac{{{x}^{2x+1}}}{{{x}^{1}}}=\left( \ln {{x}^{2x}}+2x+1 \right){{x}^{2x+1-1}}\]
\[\Rightarrow \dfrac{dy}{dx}=\left( \ln {{x}^{2x}}+2x+1 \right){{x}^{2x}}\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dx}={{x}^{2x}}\ln {{x}^{2x}}+2x\times {{x}^{2x}}+{{x}^{2x}}\]
\[\Rightarrow \dfrac{dy}{dx}={{x}^{2x}}\ln {{x}^{2x}}+2{{x}^{2x+1}}+{{x}^{2x}}\]
So, we have found the differentiation of \[y={{x}^{2x+1}}\]. The differentiation is \[\dfrac{dy}{dx}={{x}^{2x}}\ln {{x}^{2x}}+2{{x}^{2x+1}}+{{x}^{2x}}\]

Note: We should have a better knowledge in the topics of calculus and logarithms to solve this type of question easily. We should remember the following formulas to solve this type of question easily:
\[d\left( n\cdot x \right)=n\cdot dx\], where n is any constant
\[\ln {{x}^{a}}=a\ln x\]
\[\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}\]
\[d\left( \ln x \right)=\dfrac{1}{x}dx\]
\[d\left( uv \right)=v\centerdot d\left( u \right)+u\centerdot d\left( v \right)\]