Question

tell the answer ..!!!!!​

Answer 1

Answer:-

$\large\bigstar\;\underline{\boxed{\sf \dfrac{dy}{dx}=x\;e^{x}\Bigg\lgroup x+2\Bigg\rgroup}}$

Explanation:

$\rule{300}{1.5}$

Given equation,

$\longrightarrow\sf y = x^{2}\;e^{x}$

Differentiating it,

$\longrightarrow\sf \dfrac{dy}{dx}=\dfrac{d\;\bigg(x^{2}\;e^{x}\bigg)}{dx}$

Applying the product rule we get,

$\longrightarrow\sf \dfrac{dy}{dx}=\left\lgroup x^{2}\;.\;\dfrac{d\;\bigg(e^{x}\bigg)}{dx}\right\rgroup+\left\lgroup e^{x}\;.\;\dfrac{d\;\bigg(x^{2}\bigg)}{dx}\right\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}=\Bigg\lgroup x^{2}\;.\;e^{x}\Bigg\rgroup+\Bigg\lgroup e^{x}\;.\;2x \Bigg\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}= x^{2}\;e^{x}+ 2\;e^{x}\;x$

$\longrightarrow\sf \dfrac{dy}{dx}=x\;e^{x}\Bigg\lgroup x+2 \Bigg\rgroup\\\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf \dfrac{dy}{dx}=x\;e^{x}\Bigg\lgroup x+2 \Bigg\rgroup}}}}$

Hence Solved!

Formulas used:-

$\boxed{\begin{minipage}{7.5 cm} \underline{\textsf{\textbf{Formulas:}}}\\\\\bullet\;\sf\dfrac{d\;(e^{x})}{dx}=e^{x}\\\\\\\bullet\;\sf\dfrac{d\;(x^{n})}{dx}=n\;x^{n-1}\\\\\\\bullet\;\sf\dfrac{d\;[f(x)]\;[g(x)]}{dx}=f(x).\dfrac{d\;[g(x)]}{dx}+g(x).\dfrac{d\;[f(x)]}{dx}\end{minipage}}$

$\rule{300}{1.5}$

Note :- kindly see this answer from website to see the formula shown here :)

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Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathbb{SOLUTION :-}$

$❥\bf \dfrac{dy}{dx}=\dfrac{d\;\bigg(x^{2}\;e^{x}\bigg)}{dx}$

$❥ \tiny \sf \dfrac{dy}{dx}=\left\lgroup x^{2}\;.\;\dfrac{d\;\bigg(e^{x}\bigg)}{dx}\right\rgroup+\left\lgroup e^{x}\;.\;\dfrac{d\;\bigg(x^{2}\bigg)}{dx}\right\rgroup\\\\\\\\\ ❥ \sf \dfrac{dy}{dx}=\Bigg\lgroup x^{2}\;.\;e^{x}\Bigg\rgroup+\Bigg\lgroup e^{x}\;.\;2x \Bigg\rgroup\\\\\\\\ ❥ \sf \dfrac{dy}{dx}= x^{2}\;e^{x}+ 2\;e^{x}\;x$

$❥ \sf \dfrac{dy}{dx}=x\;e^{x}\Bigg\lgroup x+2 \Bigg\rgroup\\\\\\\ ❥ \large{\underline{\boxed{\blue{\bf \dfrac{dy}{dx}=x\;e^{x}\Bigg\lgroup x+2 \Bigg\rgroup}}}}$

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