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f(x)= sin √ 3x³+3x (Note its a square root of full function 3x³+3x )
Thats how I solved it.
f'(x)= Cos(3x³+3x)^½
= Cos ½(3x³+3x)^-½ × (9x²+3)
= 9x²+3Cos½ (3x^³+3x)^-½
Did I derive it right??? because solution at the back show that the answer is:
9x^2+3/2 root (3x^2+3x) cos Root (3x^3+3x)
Please Tell me if mines is right or wrong?
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Hi Gercel;
You can't just say sin(whatever) when differentiated is cos(whatever) which is what you did in the first step. To check it I need something clearer than 9x²+3Cos½ (3x^³+3x)^-½ . What does the Cos1/2 stand for. Please bracket it off better.
Last edited by bobbym (2009-07-05 07:29:54)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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First I found the derivative of the outside function Sin and Square root, by moving 1/2 to the front.. Then I did the derivatuve if the inside function 3x³+3x.
Isnt that how Chain rule should work?
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Hi Gercel;
OK, lets see, 9x²+3Cos½ (3x^³+3x)^-½ what is the first 1/2 here, please use brackets as I don't understand the answer.
I have done the problem so you can check against yours:
Last edited by bobbym (2009-07-05 08:40:25)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Wonderfull, Can you please explain how you got the answer.. please
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To do
Say:
Chain rule is:
Now we can do dy/du but du/dx is hard so we use the chain rule again:
Say:
Chain rule:
so
Now you have du/dx. Now complete the top chain rule:
Last edited by bobbym (2009-07-05 18:37:35)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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