Question

Find the derivative of 5÷3 root x

Answer 1

Step-by-step explanation:

if question is (5/3)*sqft(x) then see the photo

Answer 2

We have to find derivative of (5/3) √x

$\rm \implies \dfrac{d \bigg(\dfrac{5 \: \sqrt{x} }{3} \bigg)}{dx} \\ \\ \\ \rm \implies \dfrac{5}{3} \times \dfrac{d ({\: \sqrt{x} } )}{dx} \\ \\ \\ \rm \implies \dfrac{5}{3} \times \dfrac{d ({\: {x}^{ \frac{1}{2} } } )}{dx} \\ \\ \\ \rm \implies \dfrac{5}{3} \times \dfrac{1}{2} \times {\: {(x)}^{ \frac{1}{2} - 1} } \\ \\ \\ \rm \implies \dfrac{5}{3} \times \dfrac{1}{2} \times {\: {(x)}^{ \frac{ - 1}{2} } } \\ \\ \\ \rm \implies \dfrac{5}{6} \times \frac{1}{ {x}^{ \frac{ - 1}{2} } } \\ \\ \\ \rm \implies \dfrac{5}{6 \: \sqrt{x} }$

Additional Information :

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\ \circ \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \circ \; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\ \circ\; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \circ \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \circ \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \circ\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \circ\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\ \circ\;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\ \circ\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \circ\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \circ\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx\end{minipage}}}$