Diffentiation and integration

anpanman SG

1. Differentiate ln(x + (x^2 +1)^0.5) with respect to x.

My answer is (1/(x+(x^2+1)^0.5)) (1 + (x /+(x^2+1)^0.5))... which is the same as the answer, 1/ (x^2+1)^0.5 but I do not know how to reduce it to that form.

Next.

2 Evaluate sin x e^-cos x for the ranger of pi and 0 (integration question)

I got (-cos x (e^-cos x / - sinx)) but that seems to be wrong cuz when i subbed in the range, it's maths error.

3.Also differentiate xe^2x with respect to x.

Never seen this question anywhere in the tb before. Usually such terms only occur when we do differentiate and integrate to find the ans.

Thanks!

Forbiddensinner SG

1. Differentiate ln(x + (x^2 +1)^0.5) with respect to x.

My answer is (1/(x+(x^2+1)^0.5)) (1 + (x /+(x^2+1)^0.5))... which is the same as the answer, 1/ (x^2+1)^0.5 but I do not know how to reduce it to that form.

Next.

2 Evaluate sin x e^-cos x for the ranger of pi and 0 (integration question)

I got (-cos x (e^-cos x / - sinx)) but that seems to be wrong cuz when i subbed in the range, it's maths error.

3.Also differentiate xe^2x with respect to x.

Never seen this question anywhere in the tb before. Usually such terms only occur when we do differentiate and integrate to find the ans.

Thanks!

At one point, you should get [ 1 + (x / sqrt( x^2+1 )) ] / [ x + sqrt( x^2 + 1 ) ].

After bringing the sqrt( x^2+1 ) in the numerator down to the denominator, you should get [ x + sqrt( x^2 + 1 ) ] / [sqrt( x^2+1 )][ x + sqrt( x^2 + 1 ) ], which can be further simplified into 1 / sqrt( x^2+1 ).

When you integrate sin x e^-cos x, you should get e^-cos x.

d/dx ( xe^2x )

= x ( 2e^2x ) + e^2x ( 1 )

= e^2x ( 2x + 1 )

anpanman SG

Hi Forbiddensinner

could you explain further for question 2?

thanks.

Forbiddensinner SG

Hi Forbiddensinner

could you explain further for question 2?

thanks.

Ello,

When you differentiate cosx, you will get - sinx.

But since there is a "-" sign infront of cosx, when you bring it down, it will just become sinx.

And btw, unlike terms like 3x^5 ( d/dx = 15x^4 ) or 6x^3 ( d/dx = 18x^2 ), when you differentiate something to the power of sinx/cosx/tanx, the "power" will still remain the same.

anpanman SG

Ello,

When you differentiate cosx, you will get - sinx.

But since there is a "-" sign infront of cosx, when you bring it down, it will just become sinx.

And btw, unlike terms like 3x^5 ( d/dx = 15x^4 ) or 6x^3 ( d/dx = 18x^2 ), when you differentiate something to the power of sinx/cosx/tanx, the "power" will still remain the same.

but then this is integration...

when you integrate cos x , you'd get sin x so with the negative in front of cos x, shouldnt it be -sin x?

anpanman SG

anyway I think there's a typo for this question, could you help me check?

Show that d/dx (2x + sin 2x) = 4 cos x^2. Hence or otherwise, evaluate cos^2 x dx between pi/2 and pi/2. Ah well, since it''s between the same range, shouldnt answer be 0? However the "äctual" answer is 0.785.

next, a curve is such that d^2y/dx^2 = 6x - 2. The gradient of the curve at the point (2,-9) is 3.

Show that the gradient of the curve is never less than -16/3.

Thanks

marcteng SG

Ello,

When you differentiate cosx, you will get - sinx.

But since there is a "-" sign infront of cosx, when you bring it down, it will just become sinx.

And btw, unlike terms like 3x^5 ( d/dx = 15x^4 ) or 6x^3 ( d/dx = 18x^2 ), when you differentiate something to the power of sinx/cosx/tanx, the "power" will still remain the same.

r u a maths teacher or maths grad?

Forbiddensinner SG

anyway I think there's a typo for this question, could you help me check?

next, a curve is such that d^2y/dx^2 = 6x - 2. The gradient of the curve at the point (2,-9) is 3.

Show that the gradient of the curve is never less than -16/3.

Thanks

d/dx (2x + sin 2x)

= 2 + 2cos2x

= 2 + 2 (2 (cos x)^2 - 1)

= 2 + 4(cosx)^2 - 2

= 4(cosx)^2

I think it is -pi/2 to pi/2 for the 2nd part.

For the last part, find dy/dx, which will be 3x^2 - 2x + c.

Since they already told you that x =2 when the gradient is 3, you will find that c = -5.

Hence, dy/dx = 3x^2 - 2x - 5.

For the minimum gradient, let d^2y/dx^2 = 6x - 2 be equals to zero.

You will get x =1/3.

Sub that into dy/dx, and you will get -16/3, which is the minimum possible gradient.

Forbiddensinner SG

but then this is integration...

when you integrate cos x , you'd get sin x so with the negative in front of cos x, shouldnt it be -sin x?

But you are not integrating cosx now, technically speaking.

You are integrating sinx.

Edit: I am going out to fetch a tutee of mine soon. I will be back online while teaching if there isn't much for me to teach today.

Forbiddensinner SG

r u a maths teacher or maths grad?

NIE Undergraduate.

anpanman SG

At one point, you should get [ 1 + (x / sqrt( x^2+1 )) ] / [ x + sqrt( x^2 + 1 ) ].

When you integrate sin x e^-cos x, you should get e^-cos x.

d/dx ( xe^2x )

= x ( 2e^2x ) + e^2x ( 1 )

= e^2x ( 2x + 1 )

hi could you again explain 1 and 2 for me? I still can't manage to solve it.

by the way, for question 3, i can't solve its 2nd part which tells me to show that the integration of 4xe^2x dx = 2xe^2x - e^2x + c

This question is the reverse of what we usually do. Because the first part already tells us to differentiate xe^2x and now it wants us to integrate the term with a "xe^2x" in it. I dont see how we can make use of the 1st part...

Forbiddensinner SG

hi could you again explain 1 and 2 for me? I still can't manage to solve it.

by the way, for question 3, i can't solve its 2nd part which tells me to show that the integration of 4xe^2x dx = 2xe^2x - e^2x + c

For 1), show me your workings and I will comment on it.

For 2), I believe there is a list of terms and results for you to remember?

Such as d/dx e^sinx = cosx (e^sinx) and d/dx e^cosx = -sinx (e^cosx) ?

For 3),

4xe^2x dx

= 2[2xe^2x] dx

= 2[2xe^2x + e^2x - e^2x] dx

= [ 2[e^2x (2x+1)] - 2e^2x ] dx

I believe you can see it from here?

marcteng SG

NIE Undergraduate.

what u studying at NIE? r u good in maths in school? bet u are