Question

find the derivative of sin square x + e square into x square /logx​

Answer 1

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Question : Find the derivative of sin²x+e²x²/logx

$\huge\mathfrak{\underline{\underline{\red{Answer:-}}}}$

$\: \: \implies \displaystyle \bf{f(x) = \frac{ {sinx}^{2} + {e}^{2} \times {x}^{2} }{logx} }$

$\: \: \bullet \boxed { \sf{ \: by \: using \: quotient \: rule}}$

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{logx \frac{dy}{dx}(sin {x}^{2} + {e}^{2} {x}^{2} ) - ( \frac{dy}{dx} logx)( {sinx}^{2} + {e}^{2} {x}^{2} )}{ ({logx})^{2} } }$

$\\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{logx( \frac{dy}{dx} {sinx}^{2} + \frac{dy}{dx} {e}^{2} {x}^{2}) - ( \frac{dy}{dx} logx)( {sinx}^{2} + {e}^{2} {x}^{2}) }{ {logx}^{2} } }$

$\bullet \underline{ \boxed{ \bf{additional \: information}}}$

$\: \star \bf{ \frac{dy}{dx} {sinx}^{2} = 2sinx\times cosx.....(by \: chain \: rule})$

$\: \star \bf{ \frac{dy}{dx} {e}^{2} {x}^{2} = {e}^{2} 2x + 2 {e}^{2} {x}^{2} }........(by \: chain \: rule)$

$\: \star \bf{ \frac{dy}{dx}logx = \frac{1}{x} }$

Apply the formulas in this equation :

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{logx(2sinxcosx + {e}^{2}2x + 2e {x}^{2}) - \frac{1}{x}( {sinx}^{2} + {e}^{2} {x}^{2} ) }{ {logx}^{2} } }$

$\\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{logx \times 2sinxcosx + logx \: \times {e}^{2} {x}^{2} - \frac{1}{x} ( {sin}^{2}x + {e}^{2} {x}^{2}) }{ {logx}^{2} } }$

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{(2sinxcosxlogx + logx {e}^{2} {x}^{2} - \frac{1}{x} {sin}^{2}x - \frac{1}{e} {e}^{2} x) }{ {log}^{2}x } }$

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{(2sinxcosxlogx + logx {e}^{2} {x}^{2} - \frac{1}{x} {sin}^{2}x - {e}^{2} x) }{ {log}^{2}x } }$

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{(2sinxcosxlogx - \frac{1}{x} {sin}^{2}x + logx {e}^{2} {x}^{2} - {e}^{2}x }{ {log}^{2}x } }$

$\\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{sinx(2cosxlogx - \frac{sinx}{x} ) + {e}^{2} x(logxx - 1) }{ {log}^{2}x } }$

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{sinx( \frac{2cosxlogx x - sinx}{x}) + {e}^{2}x( logxx - 1) }{ {log}^{2}x } }$

$\: \\ \longrightarrow \displaystyle \sf{ \frac{dy}{dx} = \frac{sin(2cosxlogxx - sinx) + {e}^{2} x(logxx - 1)}{x {log}^{2}x} }$

$\: \\ \bullet \boxed{\displaystyle\sf{\dfrac{dy}{dx}=\frac{sin(2cosxlogxx-sinx) +{e}^{2} x(logxx-1)}{x{log}^{2} x} }}$