What is the derivative of $ y = {x^{5x}}? $
Answer
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Hint: As we can see that we have to find the derivative of the given question. Here the basic concept that we are going to use is the chain rule. We have to take the derivative of the composite functions and then we chain together their derivatives. The formula that we are going to use for chain rule differentiation is $ \dfrac{{df(u)}}{{dx}} = \dfrac{{df}}{{du}} \times \dfrac{{du}}{{dx}} $ .
Complete step by step solution:
We have to find the derivative of $ y = {x^{5x}}. $
First we apply the basic exponent rule which is $ {a^b} $ can be written as $ {e^{b\ln (a)}} $ . So by applying this we can write $ {x^{5x}} = {e^{5x\ln (x)}} $ , because here $ a = x $ (base).
Now we have to find the derivative of the above expression i.e.
$ \dfrac{d}{{dx}}({e^{5x\ln (x)}}) $ .
Now we will apply the chain rule
$ \dfrac{{df(u)}}{{dx}} = \dfrac{{df}}{{du}} \cdot \dfrac{{du}}{{dx}} $ .
Let us assume $ 5x\ln (x) = u $ , so by substituting this in the formula we can write $ \dfrac{d}{{du}}({e^u})\dfrac{d}{{dx}}(5x\ln (x)) $ .
We know that
$ \dfrac{d}{{du}}({e^u}) = {e^u} $ and $ \dfrac{d}{{dx}}(5x\ln (x)) $ can be written as $ 5\left( {\ln (x) + 1} \right) $ .
So we have $ \dfrac{d}{{dx}}(5x\ln (x)) = 5\left( {\ln (x) + 1} \right) $ .
We can write the new equation by putting the values together as
$ \dfrac{d}{{dx}}({x^{5x}}) = {e^u} + 5\left( {\ln (x) + 1} \right) $
Since $ \dfrac{d}{{du}}({e^u}) = {e^u} $ i.e. $ {x^{5x}} $ , so we can write $ {x^{5x}} + 5\left( {\ln (x) + 1} \right) $ .
Hence the required value is $ 5{x^{5x}}\left( {\ln (x) + 1} \right) $ .
So, the correct answer is “ $ 5{x^{5x}}\left( {\ln (x) + 1} \right) $ ”.
Note: We should note that the chain rule is very important as we have to use it in many of the problems to find the derivative of the functions. There is an alternate way to solve the above problem also by applying the product which says that $ \dfrac{d}{{dx}}\left[ {f(x)g(x)} \right] = f(x)\dfrac{d}{{dx}}[g(x)] + g(x)\dfrac{d}{{dx}}[f(x)] $ . We will first differentiate the both sides of the equation and then solve it.
Complete step by step solution:
We have to find the derivative of $ y = {x^{5x}}. $
First we apply the basic exponent rule which is $ {a^b} $ can be written as $ {e^{b\ln (a)}} $ . So by applying this we can write $ {x^{5x}} = {e^{5x\ln (x)}} $ , because here $ a = x $ (base).
Now we have to find the derivative of the above expression i.e.
$ \dfrac{d}{{dx}}({e^{5x\ln (x)}}) $ .
Now we will apply the chain rule
$ \dfrac{{df(u)}}{{dx}} = \dfrac{{df}}{{du}} \cdot \dfrac{{du}}{{dx}} $ .
Let us assume $ 5x\ln (x) = u $ , so by substituting this in the formula we can write $ \dfrac{d}{{du}}({e^u})\dfrac{d}{{dx}}(5x\ln (x)) $ .
We know that
$ \dfrac{d}{{du}}({e^u}) = {e^u} $ and $ \dfrac{d}{{dx}}(5x\ln (x)) $ can be written as $ 5\left( {\ln (x) + 1} \right) $ .
So we have $ \dfrac{d}{{dx}}(5x\ln (x)) = 5\left( {\ln (x) + 1} \right) $ .
We can write the new equation by putting the values together as
$ \dfrac{d}{{dx}}({x^{5x}}) = {e^u} + 5\left( {\ln (x) + 1} \right) $
Since $ \dfrac{d}{{du}}({e^u}) = {e^u} $ i.e. $ {x^{5x}} $ , so we can write $ {x^{5x}} + 5\left( {\ln (x) + 1} \right) $ .
Hence the required value is $ 5{x^{5x}}\left( {\ln (x) + 1} \right) $ .
So, the correct answer is “ $ 5{x^{5x}}\left( {\ln (x) + 1} \right) $ ”.
Note: We should note that the chain rule is very important as we have to use it in many of the problems to find the derivative of the functions. There is an alternate way to solve the above problem also by applying the product which says that $ \dfrac{d}{{dx}}\left[ {f(x)g(x)} \right] = f(x)\dfrac{d}{{dx}}[g(x)] + g(x)\dfrac{d}{{dx}}[f(x)] $ . We will first differentiate the both sides of the equation and then solve it.
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