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Chain Rule

Hi,

Just revising for Maths IB SL exam in two weeks and am going through the calculus section of the syllabus. While practicing I came across the following question:

x*(square root of)(3x^2 + 5)

I guess I am supposed to use the chain rule, but the final answer is supposed to be:

(3x^(1/2) + 5)^(-1/2) * (6x^2 + 5)

Could someone please explain how I am supposed to get there?

Thanks!
Original post by jack.hutch
Hi,

Just revising for Maths IB SL exam in two weeks and am going through the calculus section of the syllabus. While practicing I came across the following question:

x*(square root of)(3x^2 + 5)

I guess I am supposed to use the chain rule, but the final answer is supposed to be:

(3x^(1/2) + 5)^(-1/2) * (6x^2 + 5)

]Could someone please explain how I am supposed to get there?

Thanks!


Asuming the question is x3x2+5x\sqrt{3x^2+5} Apply the Product Rule first & then the Chain Rule!
(edited 8 years ago)
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Avatar for jack.hutch
jack.hutch
OP
2
Original post by brianeverit
Product Rule & Chain Rule!


Oh really? I would never have guessed that :s could you elaborate please?
Reply 3
Avatar for davros
davros
16
Original post by jack.hutch
Hi,

Just revising for Maths IB SL exam in two weeks and am going through the calculus section of the syllabus. While practicing I came across the following question:

x*(square root of)(3x^2 + 5)

I guess I am supposed to use the chain rule, but the final answer is supposed to be:

(3x^(1/2) + 5)^(-1/2) * (6x^2 + 5)

Could someone please explain how I am supposed to get there?

Thanks!


You need the product rule AND the chain rule, AND THEN put everything over a common denominator (or take out a common factor of (3x2+5)1/2(3x^2 + 5)^{1/2})!

Post your working if you're still stuck :smile:
Reply 4
Avatar for jack.hutch
jack.hutch
OP
2
Original post by davros
You need the product rule AND the chain rule, AND THEN put everything over a common denominator (or take out a common factor of (3x2+5)1/2(3x^2 + 5)^{1/2})!

Post your working if you're still stuck :smile:


Ok thanks for your help! I'll watch out for that trick next time it comes up.

Also sorry but I am completely stuck on another question, any chance you could give me some pointers on how to solve this type of question as well??

sinxcos(x^3)

When working out i did:

sinxcos(x^3)

= sinx - 3sinx^2 + cos(x^3) (cosx)

= cos(x^4) - 3sin(x^2) (sinx)

But apparently the answer is cos(x^4) - 3 sin (x^2) (cos(x^2))

I can't work out how to get that extra cos x^2 on the end there
Reply 5
Avatar for davros
davros
16
Original post by jack.hutch
Ok thanks for your help! I'll watch out for that trick next time it comes up.

Also sorry but I am completely stuck on another question, any chance you could give me some pointers on how to solve this type of question as well??

sinxcos(x^3)

When working out i did:

sinxcos(x^3)

= sinx - 3sinx^2 + cos(x^3) (cosx)

= cos(x^4) - 3sin(x^2) (sinx)

But apparently the answer is cos(x^4) - 3 sin (x^2) (cos(x^2))

I can't work out how to get that extra cos x^2 on the end there


I don't understand what you're trying to do there! Neither your working nor the 'answer' you've quoted make any sense.

Are you trying to differentiate sin⁡xcos⁡(x3)\sin x \cos(x^3) or sin⁡xcos⁡3(x)\sin x \cos^3(x) or something different?
Original post by jack.hutch
Ok thanks for your help! I'll watch out for that trick next time it comes up.

Also sorry but I am completely stuck on another question, any chance you could give me some pointers on how to solve this type of question as well??

sinxcos(x^3)

When working out i did:

sinxcos(x^3)

= sinx - 3sinx^2 + cos(x^3) (cosx)

= cos(x^4) - 3sin(x^2) (sinx)

But apparently the answer is cos(x^4) - 3 sin (x^2) (cos(x^2))

I can't work out how to get that extra cos x^2 on the end there


Product rule again using fact that ddx(cos3x)=−3cos⁡2xsin⁡x\frac{\text{d}}{\text{d}x}(cos^3x)=-3\cos^2x\sin x

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