Question

$find \: \frac{dy}{dx} \: if( {x +}^{2} + 2x + 1) log_{x}$
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Answer 1

Correct Question:-

Find dy/dx if y = (x² + 2x + 1) log x .

Answer:-

We have to evaluate;

$\implies \sf \: \dfrac{d}{dx} ( {x}^{2} + 2x + 1) log x$

Using d/dx (uv) = v * du/dx + u * dv/dx we get,

$\implies \sf \: log \: x \times \dfrac{d}{dx} ( {x}^{2} + 2x + 1) + ( {x}^{2} + 2x + 1) \times \dfrac{d}{dx} ( log \: x)$

Using d/dx (log x) = 1/x and d/dx (u ± v) = du/dx ± dv/dx we get,

$\implies \sf \: log \: x \times \bigg( \dfrac{d}{dx} ( {x}^{2} ) + \dfrac{d}{dx} (2x) + \dfrac{d}{dx} (1) \bigg) + ( {x}^{2} + 2x + 1) \times \dfrac{1}{x}$

Using d/dx (xⁿ) = n * x ⁿ⁻¹ & d/dx (constant) = 0 we get,

$\implies \sf \: log \: x(2 \times {x}^{2 - 1} + 1 \times 2 \times {x}^{1 - 1} + 0) + \dfrac{ {x}^{2} + 2x + 1 }{x} \\ \\ \\ \implies \sf \: log \: x(2x + 2) + \frac{ {x}^{2} + 2x + 1}{x} \\ \\ \\ \implies \sf \frac{log \: x \times x(2x + 2) + {x}^{2} + 2x + 1 }{x} \\ \\ \\ \implies \sf \underline{ \underline{ \frac{log \: x(2 {x}^{2} + 2x) + {x}^{2} + 2x + 1}{x} }}$