How do you find the derivation of $f(x) = 4\sin x + 4{x^x}$?
Answer
597k+ views
Hint: According to the question we have to determine the derivation of the given function which is as given in the question is $f(x) = 4\sin x + 4{x^x}$. So, to determine the derivation of the given function first of all we have to apply the derivation in both hand sides of the expression.
Now, we have to determine the derivation of the function which is $4\sin x$ but before finding the derivation we have to take out the constant term which is 4 as in the function and to determine the derivation we have to use the formula which is as mentioned below:
Formula used:
$ \Rightarrow \dfrac{{d\sin x}}{{dx}} = \cos x.............(A)$
Now, we have to find the derivation of the term $4{x^x}$ but same as the previous step we have to take 4 out if the derivation which is a constant term and know to determine the derivation of ${x^x}$ we can not use the formula which is as mentioned below:
$ \Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ because, on applying the formula we will get the derivation of ${x^x}$ as,
$ \Rightarrow \dfrac{{d{x^x}}}{{dx}} \ne x({x^{x - 1}})$ which is not possible.
Now, to determine the derivation of $\dfrac{{d{x^x}}}{{dx}}$ we can use the following rule which is as mentioned as ${f^g} = {e^{g\log f}}$ where, f denotes $f(x)$ and g denotes $g(x)$ and log is the natural logarithm.
Now, as we know that it is possible to take the derivation of the function easily and which is because of the derivation of ${e^x}$ which is equal to ${e^x}$ so, we only have to multiply through by the derivative of what it is given in the power.
Complete step-by-step answer:
Step 1: To determine the derivation of the given function first of all we have to apply the derivation in the both hand sides in the expression as mentioned in the solution hint. Hence,
\[ \Rightarrow \dfrac{{df(x)}}{{dx}} = \dfrac{{d(4\sin x + 4{x^x})}}{{dx}}........(1)\]
Step 2: Now, we have to determine the derivation of the function which is $4\sin x$ but before finding the derivation we have to take out the constant term which is 4 as in the function and to determine the derivation we have to use the formula (A) which is as mentioned in the solution hint. Hence,
$ \Rightarrow \dfrac{{d\sin x}}{{dx}} = \cos x$
Step 3: Now, we have to find the derivation of the term $4{x^x}$ but same as the previous step we have to take 4 out if the derivation which is a constant term and know to determine the derivation of ${x^x}$ we can not use the formula which is as mentioned below:
$ \Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ because, on applying the formula we will get the derivation of ${x^x}$ as,
$ \Rightarrow \dfrac{{d{x^x}}}{{dx}} \ne x({x^{x - 1}})$ which is not possible.
Step 4: Now, to determine the derivation of $\dfrac{{d{x^x}}}{{dx}}$ we can use the following rule which is as mentioned as ${f^g} = {e^{g\log f}}$ where, f denotes $f(x)$ and g denotes $g(x)$ and log is the natural logarithm.
Now, as we know that it is possible to take the derivation of the function easily and which is because of the derivation of ${e^x}$ which is equal to ${e^x}$ so, we only have to multiply through by the derivative of what it is given in the power. Hence,
$ \Rightarrow \dfrac{{d{x^x}}}{{dx}} = \dfrac{{d({e^x}\log x)}}{{dx}}$
Hence,
$ \Rightarrow \dfrac{{d({e^x}\log x)}}{{dx}} = {e^x}\log x\left( {\dfrac{{d(x\log x)}}{{dx}}} \right)$
$ \Rightarrow \dfrac{{d(x\log x)}}{{dx}} = 1 + \log x$
Step 5: But as we can see that there is no point leaving the result like that since we have to just go back to ${x^x}$ from ${e^{x\log x}}$. Hence,
$ \Rightarrow f'(x) = 4cox + 4{x^x}(1 + \log x)$
Hence, with the help of the formula (A) we have determined the value of the derivation $f(x) = 4\sin x + 4{x^x}$ which is $ \Rightarrow f'(x) = 4cox + 4{x^x}(1 + \log x)$.
Note:
It is possible to take the derivation of the function easily and which is because of the derivation of ${e^x}$ which is equal to ${e^x}$ so, we only have to multiply through by the derivative of what it is given in the power.
To determine the derivation of the function which is $4\sin x$ but before finding the derivation we have to take out the constant term which is 4.
Now, we have to determine the derivation of the function which is $4\sin x$ but before finding the derivation we have to take out the constant term which is 4 as in the function and to determine the derivation we have to use the formula which is as mentioned below:
Formula used:
$ \Rightarrow \dfrac{{d\sin x}}{{dx}} = \cos x.............(A)$
Now, we have to find the derivation of the term $4{x^x}$ but same as the previous step we have to take 4 out if the derivation which is a constant term and know to determine the derivation of ${x^x}$ we can not use the formula which is as mentioned below:
$ \Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ because, on applying the formula we will get the derivation of ${x^x}$ as,
$ \Rightarrow \dfrac{{d{x^x}}}{{dx}} \ne x({x^{x - 1}})$ which is not possible.
Now, to determine the derivation of $\dfrac{{d{x^x}}}{{dx}}$ we can use the following rule which is as mentioned as ${f^g} = {e^{g\log f}}$ where, f denotes $f(x)$ and g denotes $g(x)$ and log is the natural logarithm.
Now, as we know that it is possible to take the derivation of the function easily and which is because of the derivation of ${e^x}$ which is equal to ${e^x}$ so, we only have to multiply through by the derivative of what it is given in the power.
Complete step-by-step answer:
Step 1: To determine the derivation of the given function first of all we have to apply the derivation in the both hand sides in the expression as mentioned in the solution hint. Hence,
\[ \Rightarrow \dfrac{{df(x)}}{{dx}} = \dfrac{{d(4\sin x + 4{x^x})}}{{dx}}........(1)\]
Step 2: Now, we have to determine the derivation of the function which is $4\sin x$ but before finding the derivation we have to take out the constant term which is 4 as in the function and to determine the derivation we have to use the formula (A) which is as mentioned in the solution hint. Hence,
$ \Rightarrow \dfrac{{d\sin x}}{{dx}} = \cos x$
Step 3: Now, we have to find the derivation of the term $4{x^x}$ but same as the previous step we have to take 4 out if the derivation which is a constant term and know to determine the derivation of ${x^x}$ we can not use the formula which is as mentioned below:
$ \Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ because, on applying the formula we will get the derivation of ${x^x}$ as,
$ \Rightarrow \dfrac{{d{x^x}}}{{dx}} \ne x({x^{x - 1}})$ which is not possible.
Step 4: Now, to determine the derivation of $\dfrac{{d{x^x}}}{{dx}}$ we can use the following rule which is as mentioned as ${f^g} = {e^{g\log f}}$ where, f denotes $f(x)$ and g denotes $g(x)$ and log is the natural logarithm.
Now, as we know that it is possible to take the derivation of the function easily and which is because of the derivation of ${e^x}$ which is equal to ${e^x}$ so, we only have to multiply through by the derivative of what it is given in the power. Hence,
$ \Rightarrow \dfrac{{d{x^x}}}{{dx}} = \dfrac{{d({e^x}\log x)}}{{dx}}$
Hence,
$ \Rightarrow \dfrac{{d({e^x}\log x)}}{{dx}} = {e^x}\log x\left( {\dfrac{{d(x\log x)}}{{dx}}} \right)$
$ \Rightarrow \dfrac{{d(x\log x)}}{{dx}} = 1 + \log x$
Step 5: But as we can see that there is no point leaving the result like that since we have to just go back to ${x^x}$ from ${e^{x\log x}}$. Hence,
$ \Rightarrow f'(x) = 4cox + 4{x^x}(1 + \log x)$
Hence, with the help of the formula (A) we have determined the value of the derivation $f(x) = 4\sin x + 4{x^x}$ which is $ \Rightarrow f'(x) = 4cox + 4{x^x}(1 + \log x)$.
Note:
It is possible to take the derivation of the function easily and which is because of the derivation of ${e^x}$ which is equal to ${e^x}$ so, we only have to multiply through by the derivative of what it is given in the power.
To determine the derivation of the function which is $4\sin x$ but before finding the derivation we have to take out the constant term which is 4.
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