Now we have:

This isn't exactly clear, so let's write it in a different way:

Or:

Or even:

Is that clearer?

]]>P.S. thanks for explaining the fractions, it makes a lot of sense.

]]>(3 / 2) / (4 / 5) = (3 / 2) * (5 / 4) = 15 / 8

Be careful writing 1/4x. It looks like (1/4)x, when in reality it's 1 / (4x).

y = 4 + 1/4x = 4 + 4^-1 * x^-1, that 4 is on the bottom as well. Of course, 4^-1 = 1/4

y = 4 + 1/4 * x^-1

Try doing that.

]]>y = 4 + 1/4x at x = -1

= 4 + 4x^-1

dy/dx = -4x^-2 = -4/x^2 = -4/1 = -4. i.e. the Gradient of the Tangent on the curve at the stated point (x = -1) is -4 and thus the Gradient of the Normal is 1/4. My book claims it is the other way round and with opposite signs (tang is -1/4 and norm is 4)!!

]]>Ricky wrote:

-1 / (1 / 4) = -4

I've never actually been taught this

]]>-x^-2 = -1 / x^2

So if x = 1/2:

-1 / (1/2)^2 = -1 / (1 / 4) = -4

]]>find the gradiaent of the tangent and the normal to the following curves at the stated point.

y = x + 1/x at x = 1/2.

dy/dx = -x^-2 + 1 = 3/4, thus the gradient of the normal is -4/3.

I can work out most of these sums, I am doing somthing wrong with the negative power...

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y = sqrt(4/x^3) = sqrt(4x^-3) = 2x^-3/2

dy/dx = -3x^-5/2

Is this right?

y = a * f(x), y' = a * f'(x)

So:

y = 3 * (x^4), where f(x)=x^4

y' = 3 * (4x^3) = 12x^3

i.e. y=ax^b dy/dx = (ba)x^b-1]]>

Thanks.]]>

4x^-5

Otherwise, they're all right.

]]>I've moved on to the topic of calculus and am having a little trouble with some differentiation questions. I understand how to do basic questions such as y=x^10 dy/dx = 10x^9.

I think these are also correct, only my book shows the answers in a different format, I'd be obliged if you could check.

y = 1/x = x^-1 dy/dx = -x^-2 (book says: -1/x^2)

y=sqrt x = x^1/2 dy/dx = 1/2x^-1/2 (book says: 1/ 2sqrt x)

y= -1/x^4 = -x^-4 dy/dx = 4x^-3 (book says: 4/x^5)

P.S. Happy New Year!

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