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Differentiation

Can someone please help answer this?
Given that
y=3x^2+6x^1/3+((2x^3-7)/(3rootx))
Find dy/dx.
Give each term in your answer in its simplified form.
Thank you.

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Reply 1

RichE

Original post by HiggsBoson

Can someone please help answer this?
Given that
y=3x^2+6x^1/3+((2x^3-7)/(3rootx))
Find dy/dx.
Give each term in your answer in its simplified form.
Thank you.

Simplify the expression for y then use the fact that derivative of

ax^n

is

nax^(n-1).

Reply 2

HiggsBoson

Original post by RichE

Simplify the expression for y then use the fact that derivative of

ax^n

is

nax^(n-1).

I have tried that and got the first two terms correct, but not the last two. Can you tell me what the simplified expression is so I can compare it to see where i'm going wrong please.

Reply 3

RichE

Original post by HiggsBoson

I have tried that and got the first two terms correct, but not the last two. Can you tell me what the simplified expression is so I can compare it to see where i'm going wrong please.

All you need to simplify it is the power rule

x^a/x^b = x^(a-b)

Reply 4

HiggsBoson

Original post by RichE

All you need to simplify it is the power rule

x^a/x^b = x^(a-b)

I got,
dy/dx= 6x^2 + 2/x^2/3 + (5x^3/2)/3 + 7/6x^3/2
I this correct?

Reply 5

RichE

Original post by HiggsBoson

I got,
dy/dx= 6x^2 + 2/x^2/3 + (5x^3/2)/3 + 7/6x^3/2
I this correct?

Last term is wrong - or I guess is right if you mean 7/(6x^(3/2))

(edited 6 years ago)

Reply 6

HiggsBoson

Original post by RichE

Last term is wrong - or I guess is right if you mean 7/(6x^(3/2))

Whats wrong with it, i cant seem to find the problem?

Reply 7

RichE

Original post by HiggsBoson

Whats wrong with it, i cant seem to find the problem?

Your answer of 7/6x^3/2 reads like

(7/6) * x^(3/2)

rather than

(7/6) * x^(-3/2)

which is what it should represent.

You're not bracketting terms properly.

Reply 8

The Scotfather

Should it not be 6x instead of 6x^2 aswell?

Reply 9

HiggsBoson

7/6x^3/2 is (7/6) * x^(-3/2) simplified, and yeah it should be 6x lol

Reply 10

HiggsBoson

This is a core 1 paper lol, isn't there an easier way?

Reply 11

kojoterrile

https://www.derivative-calculator.net

I use this website to check my answers or help me when I’m stuck. You can use it to show steps too.

Reply 12

RichE

Original post by HiggsBoson

7/6x^3/2 is (7/6) * x^(-3/2) simplified, and yeah it should be 6x lol

No it isn't. There are rules for which operations to do first and you need to use brackets if you want operations to be done in a different order.

Reply 13

thekidwhogames

You don't have to use quotient rule. The latter fraction can be split into multiple terms of the form ax^n.

Reply 14

The Scotfather

Original post by HiggsBoson

This is a core 1 paper lol, isn't there an easier way?

Split the fraction

Reply 15

RichE

Quotient rule is not needed for this question and makes it more complicated anyway. But as it's an AS question it's clearly not expected here.

Reply 16

RichE

Just to separate out the powers

((2x^3-7)/(3rootx)) = (2/3)x^(5/2) - (7/3)x^(-1/2)

and then differentiate

Reply 17

HiggsBoson

Original post by RichE

No it isn't. There are rules for which operations to do first and you need to use brackets if you want operations to be done in a different order.

Its called the reciprocal, to get rid of a negative power you do 1 over the term with the positive power,
2x^-3 = 1/2x^3

Reply 18

The Scotfather

Here is my go at it.

(edited 6 years ago)

Snapchat-1771423696.jpg382.7KB

Reply 19

RichE

Original post by HiggsBoson

Its called the reciprocal, to get rid of a negative power you do 1 over the term with the positive power,
2x^-3 = 1/2x^3

You're still wrong.

2x^-3 means 2/(x^3).

1/2x^3 means (1/2) (x^3)

To get what you want you should write

1/(2x^3) or (2x^3)^(-1).

Look up BIDMAS if necessary. This is pre-GCSE stuff.