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Differentiate x=5y^2 + 4y - 8 with respect to x

CandidAdmin:
Differentiate x=5y^2 + 4y - 8 with respect to x

dm

CandidAdmin:
Differentiate x=5y^2 + 4y - 8 with respect to x

Multiply through the function by d/dx.
d/dx(x=5y^2 + 4y - 8 ).
Differentiate each term then make Dy/dx the subject to get
Dy/dx=1/(10y+4).

With respect to X not respect to Y.

captainbangz:
Multiply through the function by d/dx.
d/dx(x=5y^2 + 4y - 8 ).
Differentiate each term then make Dy/dx the subject to get
Dy/dx=1/(10y+4).

Busybrain2233:
With respect to X not respect to Y.

yes Sir. Lemme see your contrary result.

Since you would be differentiating with respect to X, simply make Y the subject of the formula from the giving equation, then u differentiate with respect to X.

Busybrain2233:
With respect to X not respect to Y.

Busybrain2233:
Since you would be differentiating with respect to X, simply make Y the subject of the formula from the giving equation, then u differentiate with respect to X.

I know what you're thinking, but this is implicit differentiation sir...
Do you know that the 'y' in a function with only X as independent variables doesn't magically turn to Dy/dx?
For instance, if y=x², d/dx of y will give you Dy/dx(product rule) and d/dx of x² will give you 2x.
Therefore making Dy/dx=2x?

Which results to Dy/dx=2x?

Oh my bad, person don dey forget things small small sha

Yeah, it's implicit function;

Modified:
d/dx(x)=d/dx( 5y^2 + 4y - 8 )
(1)=(10y)dy/dx+4dy/dx-0
(1)=(10y)dy/dx+4dy/dx
1=(10y+4)dy/dx
1/(10y+4)=dy/dx
Therefore: dy/dx=1/(10y+4)

Corrected

captainbangz:
I know what you're thinking, but this is implicit differentiation sir...
Do you know that the 'y' in a function with only X as independent variables doesn't magically turn to Dy/dx?
For instance, if y=x², d/dx of y will give you Dy/dx(product rule) and d/dx of x² will give you 2x.
Therefore making Dy/dx=2x?

Which results to Dy/dx=2x?

Busybrain2233:
Oh my bad, person don dey forget things small small sha

Yeah, it's implicit function;

(x)=d/dx( 5y^2 + 4y - 8 )
(1)=(10y)dy/dx+4-0
(1)=(10y)dy/dx+4
1-4=(10y)dy/dx
-3=(10y)dy/dx
-3/10y=dy/dx
Therefore: dy/dx=-3/10y

Stand to be corrected though

(x)=d/dx( 5y^2 + 4y - 8 )
(1)=(10y)dy/dx+4dy/dx-0
(1)=(10y)dy/dx+4dy/dx
1=(10y+4)dy/dx
1/(10y+4)=dy/dx
Therefore: dy/dx=1/(10y+4)

Okay sir

captainbangz:
(x)=d/dx( 5y^2 + 4y - 8 )
(1)=(10y)dy/dx+4dy/dx-0
(1)=(10y)dy/dx+4dy/dx
1=(10y+4)dy/dx
1/(10y+4)=dy/dx
Therefore: dy/dx=1/(10y+4)