Question

Differentiate ( πx⅐ + 2x + 3 ) with respect to x.

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Answer 1

$\orange{ \boxed{\boxed{\begin{array}{cc}\rm \to \:let\\ \\ \rm \:y = \pi {x}^{ \frac{1}{7} } + 2x + 3 \\ \\ \\ \blue{ \underline{ \pink{ \sf \: \: we \: have \: to \: find : }}} \\ \\ \small{\sf \hookrightarrow \: differentiate \: the \: function \: w.r.t \: \: x = \frac{dy}{dx} } \\ \\ \\ \red{ \underline{ \sf \: solution}} \\ \\ \\ \rm \: \frac{dy}{dx} = \frac{d}{dx}(\pi {x}^{ \frac{1}{7} } + 2x + 3) \\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \to \:we \: know \: that \\ \\ \sf \hookrightarrow \: \frac{d}{dx} (u \pm \: v) = \frac{d \: u}{dx} \pm \: \frac{d \: v}{dx} \\ \\ \sf \hookrightarrow \: \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \\ \\ \sf \hookrightarrow \: \frac{d}{dx}(constant) = 0 \end{array}}}}</p><p>\\ \\ \red{ \sf \: apply \: these \: rules} \\ \\ \rm \: \frac{dy}{dx} = \frac{d}{dx} \pi {x}^{ \frac{1}{7} } + \frac{d}{dx}2x + \frac{d}{dx} 3 \\ \\ \rm = \pi \times \frac{1}{7} \times {x}^{ \frac{1}{7} - 1} + 2 \times 1 + 0 \\ \\ \rm = \frac{\pi}{7} \times {x}^{ - \frac{6}{7} } + 2 \\ \\ \rm = \frac{\pi}{7 {x}^{ \frac{6}{7} } } + 2 \\ \\ \\ \blue{ \boxed{ \therefore \rm \frac{dy}{dx} = \frac{\pi}{7 {x}^{ \frac{6}{7} } } + 2}} \end{array}}}}</p><p>$

Answer 2

Explanation:

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