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c2 Differentiation

y=3(x-4)/2x^3
y=x^3(3x+8/x) diff these i carry on getting the wrong answer :frown:

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Reply 1
Original post by Ganhad
y=3(x-4)/2x^3
y=x^3(3x+8/x) diff these i carry on getting the wrong answer :frown:


What answer do you get?

For the first one,
y=3(x4)2x3=3x122x3=3x2x3122x3=32x26x3=32x26x3\displaystyle y=\frac{3(x-4)}{2x^3}=\frac{3x-12}{2x^3}=\frac{3x}{2x^3}-\frac{12}{2x^3}=\frac{3}{2x^2}-\frac{6}{x^3}=\frac32x^{-2}-6x^{-3}

Now differentiate it.

For the second one,
y=x3(3x+8x)=3x4+8x2 \displaystyle y=x^3\left(3x+\frac8{x}\right)=3x^4 + 8x^2
Differentiate it.
(edited 11 years ago)
Reply 2
Original post by TheStudentBoom
first one:

y= 3x-12/ 2x^3
y = 3x/2x^3 - 12/2x^3

y = 3/2x^2 - 6/x^3 => y= (3/2 x^-2) - (6x^-3)
dy/dx = -3x^-3 + 12x^-4

therefore dy/dx = -3/x^3 + 12/x^4

might have done something wrong there, but thats what I know. I'll do the second one in a seperate post.


Your answer is wrong.

By the way, please don't post full solutions. Full solutions are considered a last resort.
Original post by raheem94
What answer do you get?

For the first one,
y=3(x4)2x3=3x122x3=3x2x3122x3=32x26x3=32x26x3\displaystyle y=\frac{3(x-4)}{2x^3}=\frac{3x-12}{2x^3}=\frac{3x}{2x^3}-\frac{12}{2x^3}=\frac{3}{2x^2}-\frac{6}{x^3}=\frac32x^2-6x^{-3}



hey the last bit is 3/2 x^-2 - 6x^-3 right? not 3/2 x^2?
Original post by raheem94
Your answer is wrong.

By the way, please don't post full solutions. Full solutions are considered a last resort.


Very sorry, I'll just delete my posts then.
Reply 5
Original post by TheStudentBoom
hey the last bit is 3/2 x^-2 - 6x^-3 right? not 3/2 x^2?


Thanks for indicating my mistake. I will edit my post.
Reply 6
Original post by TheStudentBoom
Very sorry, I'll just delete my posts then.


+rep to you :smile:
Reply 7
Original post by raheem94
+rep to you :smile:


book says something different i did get what you got -_- Damn Book could you show me this one aswell y=3sqrtx(x-2)^2
Reply 8
Original post by Ganhad
book says something different i did get what you got -_- Damn Book could you show me this one aswell y=3sqrtx(x-2)^2


What is the answer in the book for the previous question?
Reply 9
Original post by Ganhad
book says something different i did get what you got -_- Damn Book could you show me this one aswell y=3sqrtx(x-2)^2


Is this your new question y=3x(x2)2 \displaystyle y=3\sqrt{x(x-2)^2}
Reply 10
no the root does not go over the bracket just over the x :wink:

Original post by raheem94
Is this your new question y=3x(x2)2 \displaystyle y=3\sqrt{x(x-2)^2}
(edited 11 years ago)
Reply 11
Original post by Ganhad
no the root does not go over the bracket just over the x :wink:


y=3x(x2)2=3x12(x24x+4)=3x5212x32+12x12 \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}

Now its simple to differentiate it.

By the way, what is the answer of the previous question in the book?
Reply 12
im supposed to find the gradient of that curve where x = 3 the answer is supposed to be 29root3 could you check if you get that please
Reply 13
Original post by raheem94
y=3x(x2)2=3x12(x24x+4)=3x5212x32+12x12 \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}

Now its simple to differentiate it.

By the way, what is the answer of the previous question in the book?


-3x^-3+6x^-4
Reply 14
Original post by Ganhad
im supposed to find the gradient of that curve where x = 3 the answer is supposed to be 29root3 could you check if you get that please


Which question is it?

You have posted 3 questions in this thread, which is this?
Reply 15
Original post by Ganhad
-3x^-3+6x^-4


Is this the answer to y=3(x4)2x3=3x122x3=3x2x3122x3=32x26x3=32x26x3 ?\displaystyle y=\frac{3(x-4)}{2x^3}=\frac{3x-12}{2x^3}=\frac{3x}{2x^3}-\frac{12}{2x^3}=\frac{3}{2x^2}-\frac{6}{x^3}=\frac32x^{-2}-6x^{-3} \ ?

If this is the case then it is wrong, the correct answer should be 3x3+18x4 \displaystyle -3x^{-3}+18x^{-4}
Reply 16
i agree but could you see if u get 29 root 3 if you sub x=3 into y=3x(x2)2=3x12(x24x+4)=3x5212x32+12x12 \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}
Reply 17
Original post by Ganhad
i agree but could you see if u get 29 root 3 if you sub x=3 into y=3x(x2)2=3x12(x24x+4)=3x5212x32+12x12 \displaystyle y=3\sqrt{x}(x-2)^2 = 3x^{\frac12}(x^2-4x+4)=3x^{\frac52}-12x^{\frac32}+12x^{\frac12}


Substituting x=3 in y=3x(x2)2 \displaystyle y=3\sqrt{x}(x-2)^2 give 33 3\sqrt3

Substituting x=3 in the derivative of y=3x(x2)2 \displaystyle y=3\sqrt{x}(x-2)^2 gives 1332 \displaystyle \frac{13\sqrt3}2
Reply 18
this book is wak -_- so many damn mistakes
Reply 19
Original post by raheem94
Substituting x=3 in y=3x(x2)2 \displaystyle y=3\sqrt{x}(x-2)^2 give 33 3\sqrt3

Substituting x=3 in the derivative of y=3x(x2)2 \displaystyle y=3\sqrt{x}(x-2)^2 gives 1332 \displaystyle \frac{13\sqrt3}2


the equation of a curve is y= x^2-5x-24/x

find the gradient of the curve at each of the point were the curve crosses the x-axis how would you do this sorry for asking for so much help but im a thicko that has been doing this for neardy 1 1/2 days straight :wink:

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