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# C3 Differentiation help Watch

1. Hi,

I've just been working through some questions and am not sure about a couple of them. I would be immensely grateful for someone to check my answers.

1. Differentiating e^(3x).(sinx +2cosx) = e^(3x).(cosx - 2 sinx) + 3e^(3x).(sinx +2cosx)
= e^(3x) (7cosx + sinx)?

2. y=(3xe^x) - 1
dy/dx = (3e^x).(x+1)
Turning point at x=-1 only as otherwise x=ln0 (which is impossible)
d2y/dx2 = (3e^x).(1+(x+1)) = (3e^x).(2+x)?

3. y=4sec5x.(sin2x)^3
dy/dx = 4(sinx)^2 . (6sec5xcos2x + 5sec5xtan5xsin2x)?

4. y=(x^3)ln(5x+2)
dy/dx = (x^2)(5x/5x+2 + 3ln(5x+2))? can this be simplified anymore?

5. y = x(1+x)^(1/2)
My answer for dy/dx = (x+3(1+x)^(1/2)) / 2(1+x)^(1/2)
Correct answer = dy/dx = (3x+2) / 2(1+x)^(1/2)

Turning point at x= -2/3
Correct answer for y = (-2.(3^1/2)) / 3.(3^1/2)

Finally, point P is (-1,0). The gradient is apparently infinite at this point - why is that?
dy/dx = -1/0 at P.

2. (Original post by PardonMyFrench96)
Hi,

I've just been working through some questions and am not sure about a couple of them. I would be immensely grateful for someone to check my answers.

1. Differentiating e^(3x).(sinx +2cosx) = e^(3x).(cosx - 2 sinx) + 3e^(3x).(sinx +2cosx)
= e^(3x) (7cosx + sinx)?

2. y=(3xe^x) - 1
dy/dx = (3e^x).(x+1)
Turning point at x=-1 only as otherwise x=ln0 (which is impossible)
d2y/dx2 = (3e^x).(1+(x+1)) = (3e^x).(2+x)?

3. y=4sec5x.(sin2x)^3
dy/dx = 4(sinx)^2 . (6sec5xcos2x + 5sec5xtan5xsin2x)?

4. y=(x^3)ln(5x+2)
dy/dx = (x^2)(5x/5x+2 + 3ln(5x+2))? can this be simplified anymore?

5. y = x(1+x)^(1/2)
My answer for dy/dx = (x+3(1+x)^(1/2)) / 2(1+x)^(1/2)
Correct answer = dy/dx = (3x+2) / 2(1+x)^(1/2)

Turning point at x= -2/3
Correct answer for y = (-2.(3^1/2)) / 3.(3^1/2)

Finally, point P is (-1,0). The gradient is apparently infinite at this point - why is that?
dy/dx = -1/0 at P.

Okay, 1-4 look completely correct to me (although obviously I may have made a mistake too). I don't think 4 can be meaningfully simplified. I'm not entirely sure how you got to your answer for dy/dx for 5. Just differentiate it normally using the product rule and then cross multiply when you're left with two terms which should lead you to the correct answer. There is indeed a turning point at x=-2/3 but I'm not entirely sure how you or the "correct answer" got to that value of y... if you sub that value of x into y, you should get -2root3/9, unless I've misunderstood you. The gradient is infinite at (-1,0). dy/dx = -1/0 which is undefined, but dx/dy = 0/-1 = 0. So if you flip the axes around, you've got a horizontal line so if you flip them back again, you've got a vertical line which therefore has an infinite gradient.
3. (Original post by Chlorophile)
Okay, 1-4 look completely correct to me (although obviously I may have made a mistake too). I don't think 4 can be meaningfully simplified. I'm not entirely sure how you got to your answer for dy/dx for 5. Just differentiate it normally using the product rule and then cross multiply when you're left with two terms which should lead you to the correct answer. There is indeed a turning point at x=-2/3 but I'm not entirely sure how you or the "correct answer" got to that value of y... if you sub that value of x into y, you should get -2root3/9, unless I've misunderstood you. The gradient is infinite at (-1,0). dy/dx = -1/0 which is undefined, but dx/dy = 0/-1 = 0. So if you flip the axes around, you've got a horizontal line so if you flip them back again, you've got a vertical line which therefore has an infinite gradient.
Thanks - for question 5 I got what you did, but the correct answer is apparently -2root3/3root3. Thanks for explaining that.
4. (Original post by PardonMyFrench96)
Thanks - for question 5 I got what you did, but the correct answer is apparently -2root3/3root3. Thanks for explaining that.

so that "answer" doesn't make much sense unless I'm misreading what you've written!

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