Question

find the derivativey=√(4x - 7)​

Answer 1

$\sf \rightarrow \quad y = \sqrt{4x - 7} \\ \\ \sf \rightarrow \quad y' = \frac{d( \sqrt{4x - 7}) }{dx} \\ \\ \sf \rightarrow \quad y' = \frac{d{(4x - 7)}^{ \frac{1}{2} } }{d(4x - 7)} \cdot \frac{d(4x - 7)}{dx} \qquad \quad \bigg[ \because \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \bigg] \\ \\ \sf \rightarrow \quad y' = \frac{1}{2} \cdot {(4x - 7)}^{ - \frac{1}{2}} \cdot \bigg(\frac{d(4x)}{dx} - \frac{d(7)}{dx} \bigg) \\ \\ \sf \rightarrow \quad y' = \frac{1}{ \cancel{2}_{1} \sqrt{4x - 7} } \times \cancel{4}^{2} \qquad \quad \big[ \because {a}^{ - m} = \frac{1}{ {a}^{m} } \big] \\ \\ \sf \rightarrow \quad y' = \frac{2}{ \sqrt{4x - 7} } \times \frac{ \sqrt{4x - 7} }{ \sqrt{4x - 7} } \\ \\ \sf \rightarrow \quad y' = \frac{2 \sqrt{4x - 7} }{4x - 7} \\ \\ \\ \underline{ \sf Some \: Formulae}- \\ \\ \star \quad \sf \frac{d(cos \: x)}{dx} = - sin \: x \\ \\ \star \quad \sf \frac{d(sin \: x)}{dx} = cos \: x \\ \\ \star \quad \sf \frac{d(ln \: x)}{dx} = \frac{1}{x} \\ \\ \star \quad \sf \frac{d(u + v)}{dx} = \frac{du}{dx} + \frac{dv}{dx} \\ \\ \star \quad \sf \frac{d(tan \: x)}{dx} = {sec}^{2} \: x \\ \\ \star \sf \quad \frac{d(cot \: x)}{dx} = - {cosec}^{2} \: x \\ \\ \star \sf \quad \frac{d( {e}^{x}) }{dx} = {e}^{x} \\ \\ \star \quad \sf \frac{da}{dx} = 0 \\ \sf \qquad where, \: a = constant$

Answer 2

$\huge \mathfrak{ \green{Solution}}$

$\sf \longrightarrow{ \purple{ \quad y = \sqrt{4x - 7}}}$

$\sf \longrightarrow{ \purple{ \quad y' = \frac{d( \sqrt{4x - 7}) }{dx}}}$

$\sf \longrightarrow { \purple{\quad y' = \frac{d{(4x - 7)}^{ \frac{1}{2} } }{d(4x - 7)} \cdot \frac{d(4x - 7)}{dx} \qquad \quad \bigg[ \because \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \bigg] }}$

$\sf \longrightarrow{ \purple{ \quad y' = \frac{1}{2} \cdot {(4x - 7)}^{ - \frac{1}{2}}\cdot\bigg(\frac{d(4x)}{dx} - \frac{d(7)}{dx} \bigg) }}$

$\sf \longrightarrow { \purple{\quad y' = \frac{1}{ \cancel{2}_{1} \sqrt{4x - 7} } \times \cancel{4}^{2} \qquad \quad \big[ \because {a}^{ - m} = \frac{1}{ {a}^{m} } \big] }}$

$\sf \longrightarrow { \purple{\quad y' = \frac{2}{ \sqrt{4x - 7} } \times\frac{ \sqrt{4x - 7} }{ \sqrt{4x - 7} }}}$

$\sf \longrightarrow { \red{\quad y' = \frac{2 \sqrt{4x - 7} }{4x - 7} }}$

$\begin{gathered}\tiny\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\\begin{gathered} \underline{ \color{navy}{{ \sf Some \: More \: Formulae \: - }}} \\ \\ \star \quad \sf \frac{d(cos \: x)}{dx} = - sin \: x \\ \\ \star \quad \sf \frac{d(sin \: x)}{dx} = cos \: x \\ \\ \star \quad \sf \frac{d(ln \: x)}{dx} = \frac{1}{x} \\ \\ \star \quad \sf \frac{d(u + v)}{dx} = \frac{du}{dx} + \frac{dv}{dx} \\ \\ \star \quad \sf \frac{d(tan \: x)}{dx} = {sec}^{2} \: x \\ \\ \star \sf \quad \frac{d(cot \: x)}{dx} = - {cosec}^{2} \: x \\ \\ \star \sf \quad \frac{d( {e}^{x}) }{dx} = {e}^{x} \\ \\ \star \quad \sf \frac{da}{dx} = 0 \\\\ \sf where, a = constant \end{gathered} {}\end{array}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$