Question

Solve this equation pls

Answer 1

Answer:

${ \rm{y = \frac{3 \sqrt{x} + 2x}{ {4x}^{2} } }}$

• We are gonna use the quotient rule

${ \boxed{ \bf{ \frac{dy}{dx} = \frac{ {\huge{v} } \frac{du}{dx} - { \huge{u}} \frac{dv}{dx} }{ {v}^{2} } }}}$

• u is (3√x + 2x)
• v is 4x²
• du/dx is 3/2√x
• dv/dx is 8x

• Therefore:

${ \rm{ \frac{dy}{dx} = \frac{(4 {x}^{2})( \frac{3}{2 \sqrt{x} }) - (3 \sqrt{x} + 2x)(8x) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{(6 {x}^{2})( {x}^{ - \frac{1}{2} } ) - (24 {x}^{ \frac{3}{2} } + 16 {x}^{2} ) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ 6 {x}^{ \frac{3}{2} } - {24x}^{ \frac{3}{2} } - {16x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ - 18 {x}^{ \frac{3}{2} } - 16 {x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = - 2 {x}^{ \frac{3}{2} } (9 + 8 {x}^{ \frac{4}{3} } ) \div 16 {x}^{4} }} \\ \\ { \boxed{ \boxed{ \rm{ \: \: \frac{dy}{dx} = - \frac{8 \sqrt{ {x}^{ \frac{8}{3} }} + 9}{8x {}^{ \frac{8}{3} } } \: \: }}}}$

Answer 2

I'll answer ur question later

Answer mine now

Do u find me annoying ?

Pls be honest

So that I won't bother u More