Question

Answer the question which is in attachment​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Let:

$\tt \implies y = \dfrac{ {x}^{5} - cos(x) }{sin(x) }$

$\tt \implies \dfrac{dy}{dx} = \dfrac{d}{dx} \bigg ( \dfrac{ {x}^{5} - cos(x) }{sin(x) } \bigg)$

Using differentiation rule:

$\tt \implies \dfrac{d}{dx} \bigg( \dfrac{f}{g} \bigg) = \dfrac{g \cdot \dfrac{d}{dx}(f) - f \cdot \dfrac{d}{dx}(g) }{ {g}^{2} }$

$\tt \implies \dfrac{dy}{dx} = \dfrac{sin(x) \cdot \dfrac{d}{dx} \big( {x}^{5} - cos(x) \big) - \big( {x}^{5} - cos(x) \big) \cdot \dfrac{d}{dx} \big(sin(x) \big) }{ {sin}^{2}(x)}$

$\tt \implies \dfrac{dy}{dx} = \dfrac{sin(x) \big(5 {x}^{4} - ( - sin(x)) \big) - \big( {x}^{5} - cos(x) \big) \times cos(x) }{ {sin}^{2}(x)}$

$\tt \implies \dfrac{dy}{dx} = \dfrac{sin(x) \big(5 {x}^{4} + sin(x) \big) - \big( {x}^{5} - cos(x) \big) \times cos(x) }{ {sin}^{2}(x)}$

$\tt \implies \dfrac{dy}{dx} = \dfrac{5 {x}^{4} \: sin(x) + sin^{2} (x) - {x}^{5} \: cos(x) + cos^{2} (x) }{ {sin}^{2}(x)}$

$\tt \implies \dfrac{dy}{dx} = \dfrac{5 {x}^{4} \: sin(x) - {x}^{5} \: cos(x) +1}{ {sin}^{2}(x)}$

Which is our required answer.

$\texttt{\textsf{\large{\underline{Learn More}:}}}$

$\tt 1. \: \dfrac{d}{dx}sin(x) = cos(x)$

$\tt 2. \: \dfrac{d}{dx}cos(x) = - sin(x)$

$\tt 3. \: \dfrac{d}{dx}tan(x) = sec^{2} (x)$

$\tt 4. \: \dfrac{d}{dx}cot(x) = - cosec^{2} (x)$

$\tt 5. \: \dfrac{d}{dx}sec(x) = sec(x) \: tan(x)$

$\tt 6. \: \dfrac{d}{dx}cosec(x) =- cosec(x) \: cot(x)$