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differentiation (1 Viewer)
- Thread starter fullonoob
- Start date
i dont get how d(x+y)/[dx.d(x+y)] changes to 1 + dy /dx in your methodin detail:
If you live in my area and need a good(????) private maths tutor, pm me. Not cheap though.
d/dx(x+y) = 1 + dy / dx, but how does the d(x+y) disappear
I think you have it all mixed up.
Chain rule my way is no different from the regular chain rule except I don't bother with the formal substituting and subsequent time-wasting "unsubstituting"
}{\mathrm{d} (x+y)}\times \frac{\mathrm{d} (x+y)}{\mathrm{d} x}\ =\ \frac{\mathrm{d} cos(x-y)}{\mathrm{d} (x-y)}\times \frac{\mathrm{d} (x-y)}{\mathrm{d} x}\\ \\ $ On the LHS: $ \ \ \frac{\mathrm{d} sin(x+y)}{\mathrm{d}(x+y)}\ = cos(x+y)\ \ $just as $\\ \\ \frac{\mathrm{d} sin\theta }{\mathrm{d} \theta }\ =\ cos \theta \ \ \ and\ \ \ \frac{\mathrm{d}\ sin(donkey)}{\mathrm{d}\ donkey}\ =\ cos(donkey)\\ \\ and\ \ \frac{\mathrm{d} sin\ apple}{\mathrm{d} \ apple} \ =\ cos\ apple\ \ \ and\ \ \ \frac{\mathrm{d} \ sin(3xy^{2}-\frac{y}{x})}{\mathrm{d} (3xy^{2}-\frac{y}{x})}\ =\ cos(3xy^{2}-\frac{y}{x})\\ \\ \\ As\ for\ the\ 2nd\ part:\ \ \frac{\mathrm{d} (x+y)}{\mathrm{d} x}\ =\ \frac{\mathrm{d} x}{\mathrm{d} x}+\frac{\mathrm{d} y}{\mathrm{d} x}\ =\ 1+\frac{\mathrm{d} y}{\mathrm{d} x} )
I hope it is clearer now.
I tutor maths 1-on-1 in the Epping/Beecroft/Eastwood area. Maybe you'll find me quite good.
Chain rule my way is no different from the regular chain rule except I don't bother with the formal substituting and subsequent time-wasting "unsubstituting"
I hope it is clearer now.
I tutor maths 1-on-1 in the Epping/Beecroft/Eastwood area. Maybe you'll find me quite good.
Last edited:
oh i get it now thanks.
Help me on another question
y = 2sinx + cos2x
find the x intercepts
0 =2sinx + cos2x
0 = 2sinx + 1-2sin^2x
2sin^2x - 2sinx-1 = 0
let u = sinx
2u^2 - 2u -1 = 0
delta = 4-4(2)(-1) = 12
root delta = 2 root 3
u = (2 +- 2 root 3)/4
sinx = (1+-root3)/ 2
x= sin^-1 (1+-root3)/ 2
only one of the cases work i.e. -21.47 degrees (cbb typing the repeater for this degree)
how come i cant get the other x intecept, using domain -2pi <x<2pi
Help me on another question
y = 2sinx + cos2x
find the x intercepts
0 =2sinx + cos2x
0 = 2sinx + 1-2sin^2x
2sin^2x - 2sinx-1 = 0
let u = sinx
2u^2 - 2u -1 = 0
delta = 4-4(2)(-1) = 12
root delta = 2 root 3
u = (2 +- 2 root 3)/4
sinx = (1+-root3)/ 2
x= sin^-1 (1+-root3)/ 2
only one of the cases work i.e. -21.47 degrees (cbb typing the repeater for this degree)
how come i cant get the other x intecept, using domain -2pi <x<2pi