Respuesta :

remotecontrolbrain

Answer: [tex]\frac{dy}{dx}=\sqrt{20x^3-4x}[/tex]

Step by step:

For better understanding I am using a new variable z instead of original x and then substitute x^2 for z. This makes the use of the chain rule clearer. Let me know if you have questions.

[tex]y=f(z)\\f'(z)=\frac{dy}{dz}=\sqrt{5z-1}\\\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}\\z=x^2\\y=f(x^2)\\\frac{dy}{dx}=f'(x^2)\frac{d(x^2)}{dx}\\\frac{dy}{dx}=\sqrt{5x^2-1}\cdot2x\\\frac{dy}{dx}=\sqrt{20x^3-4x}[/tex]

apologiabiology

use chain rule

remember that [tex]\frac{d}{dx} g(h(x))=g'(h(x))h'(x)[/tex] (mixing Lagrange's notation and Leibniz's notation)


[tex]\frac{dy}{dx}=[/tex]

[tex]\frac{d}{dx}y=[/tex]

[tex]\frac{d}{dx}f(x^2)=[/tex]


if we treat f(x) as g(x) and [tex]x^2[/tex] as h(x) then setup parts

[tex]f'(x)=\sqrt{5x-1}[/tex] (given) so [tex]f'(x^2)=\sqrt{5x^2-1}[/tex]

and h'(x)=[tex]\frac{d}{dx}x^2=2x[/tex]

[tex]\frac{d}{dx}y=[/tex]

[tex](\sqrt{5x^2-1})(2x)=[/tex]

[tex]2x\sqrt{5x^2-1}[/tex]


answer: [tex] \frac{dy}{dx}=2x\sqrt{5x^2-1}[/tex]