Substitution Question Query. (1 Viewer)

Prashant Sallan · Oct 8, 2009

This is for Mx1 + Mx2, substitution questions:

When the substitution is give as:

y = sqrt (2x - 3)

Are we allowed to rearrange the substitution to:

y^2 = (2x - 3)

(and then ddx both sides) ?

Thanks.

the-derivative · Oct 8, 2009

Prashant Sallan said:
This is for Mx1 + Mx2, substitution questions:

When the substitution is give as:

y = sqrt (2x - 3)

Are we allowed to rearrange the substitution to:

y^2 = (2x - 3)

(and then ddx both sides) ?

Thanks.

I would think in the HSC yes, but fundamentally you ARE using their substitution, but instead of differentiating initially, you're squaring first.

The only concern I'd have is whether implicit differentiation would be awarded marks in an Extension I exam on the criteria (should be fine in Extension II).

Drongoski · Oct 8, 2009

Do you mean:

$\100dpi \frac{\mathrm{d} }{\mathrm{d} x}\sqrt{2x-3}\ \ ?\\ \\ y\ =\ \sqrt{2x-3}\ =\ (2x-3)^{\frac{1}{2}}\\ \\ \therefore \ \ \frac{\mathrm{d} y}{\mathrm{d} x}\ =\ \frac{1}{2}(2x-3)^{-\frac{1}{2}}\cdot 2\ =\ \frac{1}{\sqrt{2x-3}}\\ \\ y^{2}\ =\ (\sqrt{2x-3})^{2}\ =\ 2x-3\\ \\ \frac{\mathrm{d} }{\mathrm{d} x}\ both\ sides:\ \ \ 2y\frac{\mathrm{d} y}{\mathrm{d} x}\ =\ 2\\ \\ \therefore \ \ \frac{\mathrm{d} y}{\mathrm{d} x}\ =\ \frac{2}{2y}\ =\ \frac{1}{\sqrt{2x-3}}\\ \\ \\$

Same results

kwabon · Oct 8, 2009

Prashant Sallan said:
This is for Mx1 + Mx2, substitution questions:

When the substitution is give as:

y = sqrt (2x - 3)

Are we allowed to rearrange the substitution to:

y^2 = (2x - 3)

(and then ddx both sides) ?

Thanks.

can use either of the two methods, HSC markers accept both.

(y)

Substitution Question Query. (1 Viewer)

Prashant Sallan

Member

the-derivative

BCom/LLB (UNSW)

Drongoski

Well-Known Member

kwabon

Banned

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