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**Hint:**It must be remembered that \[\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x} \times \ln (a)\] and also how we do a derivative when u and v both are functions in x and they both are multiplied. Thus, \[\dfrac{d}{{dx}}\left( {u \times v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] .

**Complete step-by-step answer:**As we know previously that \[\dfrac{d}{{dx}}\left( {u \times v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] So here we will consider \[u = {5^x}\& v = {x^5}\]

So now it becomes

\[\begin{array}{l}

\therefore y = {5^x}{x^5}\\

\Rightarrow \dfrac{{dy}}{{dx}} = {5^x}\dfrac{d}{{dx}}{x^5} + {x^5}\dfrac{d}{{dx}}{5^x}\\

\Rightarrow \dfrac{{dy}}{{dx}} = {5^x} \times 5{x^4} + {x^5} \times {5^x} \times \log (5)

\end{array}\]

This is because \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\& \dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x} \times \ln (a)\]

So now, further solving it, we are getting

\[ \Rightarrow \dfrac{{dy}}{{dx}} = {5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)\]

**Which means that option D is the correct option here.**

**Note:**A lot of students will take log on both sides as their very preliminary stage but that is not necessary and it will only just complicate things, yes we take logarithms on both sides to bring the power to bases and we are also aware of the fact that derivative of logarithmic y will be the inverse of y. But Using that here will Stretch the solution far more lengthier and thus chances of making silly mistakes will also be higher.

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