Question

find dy/dx for y=sq root (2x+1) (x^3)
plz show how to solve plz and ty!!!!!

Answer 1

i think ur math is \[y=\sqrt{(2x+1)x ^{3}}\]

Answer 2

\(y=\sqrt{x^3(2x+1)}\)?

Answer 3

i hv to find out\[\frac{ dy }{ dx }\]

Answer 4

sq root (2x+1) then (x^3)
(x^3 is not under sq root)

Answer 5

ok

Answer 6

\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 7

\(y=x^3\sqrt{2x+1}=\sqrt{2x^7+x^6}\)$$\frac{dy}{dx}=\frac{14x^6+6x^5}{\sqrt{2x^7+x^6}}$$

Answer 8

\[\sqrt{2x+1}x^2\] use
\[(fg)'=f'g+g'f\] with
\[f(x)=\sqrt{2x+1},f'(x)=\frac{1}{\sqrt{x+1}},g(x)=x^3,g'(x)=3x^2\]

Answer 9

$$\frac{14x^6+6x^5}{2x^3\sqrt{2x+1}}=\frac{7x^3+3x^2}{\sqrt{2x+1}}=\frac{x^2(7x+3)\sqrt{2x+1}}{2x+1}$$

Answer 10

It's also sometimes helpful to try logarithmic differentiation with problems like these.

Answer 11

ya

Answer 12

let\[\ln (y)=\ln(\sqrt{2x+1}x ^{3})=\ln \sqrt{2x+1}+lnx ^{3}=\frac{ 1 }{ 2 }\ln(2x+1)+3lnx\]

Answer 13

now u hv to differentiate both sides

Answer 14

if u need further steps plz ask me