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C3 Differentiation

jasminetwine

Hi guys

Is anybody able to tell me what the happens in part b in the last step of the model answer?

Thanks 😄

Reply 1

NotNotBatman

Common denominator in the square root.

Reply 2

Zacken

Original post by jasminetwine

Hi guys

Is anybody able to tell me what the happens in part b in the last step of the model answer?

Thanks 😄

Specifically:

1 - \frac{x^2}{16} = \frac{16 - x^2}{16}

\sqrt{1 - \frac{x^2}{16}} = \sqrt{\frac{16 - x^2}{16}} = \frac{\sqrt{16-x^2}}{\sqrt{16}} = \frac{\sqrt{16-x^2}}{4}

.

Hence:

8\sqrt{1 - \frac{x^2}{16}} = 8 \times \frac{\sqrt{16-x^2}}{4} = 2\sqrt{16-x^2}

GLORIOUS LATEX.

Reply 3

jasminetwine

Original post by Zacken

Specifically:

1 - \frac{x^2}{16} = \frac{16 - x^2}{16}

\sqrt{1 - \frac{x^2}{16}} = \sqrt{\frac{16 - x^2}{16}} = \frac{\sqrt{16-x^2}}{\sqrt{16}} = \frac{\sqrt{16-x^2}}{4}

.

Hence:

8\sqrt{1 - \frac{x^2}{16}} = 8 \times \frac{\sqrt{16-x^2}}{4} = 2\sqrt{16-x^2}

GLORIOUS LATEX.

Thank you very much Zacken :smile:

I'm not sure I would have ever spotted that!

Do you think a student would still be able to achieve full marks if the answer was left unsimplified as dy/dx=+- 1/(8(1-1/16x^2)) (i.e. The penultimate line)?

Reply 4

Zacken

Original post by jasminetwine

Thank you very much Zacken :smile:

I'm not entirely sure, but unless asked for a specific form, I don't think they should lose any marks.

Reply 5

Kallisto

Entertainment Forum Helper

Original post by NotNotBatman

Common denominator in the square root.

That is right! you have razor-sharp eyes for understanding mathematical steps. :congrats:

Original post by jasminetwine

Hi guys

Is anybody able to tell me what the happens in part b in the last step of the model answer?

Thanks 😄

What about the steps before? did you understand them? well.

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