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# If a function is given as $y = {5^x}{x^5}$ , then $\dfrac{{dy}}{{dx}}$ is A).${5^x}\left( {x\log 5 - 5{x^4}} \right)$B). ${{x^5}\log 5 - 5{x^4}}$C). ${{x^5}\log 5 + 5{x^4}}$D). ${5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)$

Last updated date: 23rd Jul 2024
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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, $\dfrac{d}{{dx}}\left( {u \times v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ .

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 $u = {5^x}\& v = {x^5}$
$\begin{array}{l} \therefore y = {5^x}{x^5}\\ \Rightarrow \dfrac{{dy}}{{dx}} = {5^x}\dfrac{d}{{dx}}{x^5} + {x^5}\dfrac{d}{{dx}}{5^x}\\ \Rightarrow \dfrac{{dy}}{{dx}} = {5^x} \times 5{x^4} + {x^5} \times {5^x} \times \log (5) \end{array}$
This is because $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\& \dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x} \times \ln (a)$
$\Rightarrow \dfrac{{dy}}{{dx}} = {5^x}\left( {{x^5}\log 5 + 5{x^4}} \right)$