How does the derivative of y^2 with respect to x is 2yy' I get the 2y part but not the y'?

[flint ravine]

Flames

[solar loom]

when you take the derivative of $f(x)=x^2$ what you get is $f'(x)=2x\frac{dx}{dx}=2xx'$ but since x' is just 1, you ignore it. $\$ when you take the derivative of $f(x)=y^2$ you're doing the same thing, but this time the y' term doesn't just equal 1, so you have to include it

[flint ravine]

Robert

[solar loom]

this is all assuming you're differentiating with respect to x, which is safe to assume here

[eager violet]

so basically since x' is dx/dx, we can divide that to 1 and dy/dx since there are different we keep it as dy/dx or y'??

[solar loom]

yes

[eager violet]

alr thank you @solar loom

how to close ticket

[solar loom]

.close

[eager violet]

.close