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akhl

@akhl

Dazzler

Posted: 15 years ago

#161

Find y ' and y ".

y=ln(4x)/x^2

y = ln(4x)/x^2 -----------------(1)

x^2 * y = ln(4x)

Differentiating both sides with respect to x,

x^2 * y' + 2xy = 1/(4x) * 4

x^2 * y' + 2xy = 1/x

Dividing throughout by x^2,

y' + 2y/x = 1/x^3

y' = 1/x^3 - 2y/x

y' = 1/x^3 - 2ln(4x)/x^3

y' = (1 - 2 ln(4x))/x^3-------------(2)

Or x^3 y' = 1 - 2 ln(4x)

Differentiating both sides with respect to x,

x^3 y ' ' + 3x^2 y' = -2 * 1/(4x) * 4

x^3 y' ' + 3x^2y' = - 2/x

Dividing throughout by x^3

y ' ' + 3 y'/x = -2/x^4

y ' ' = -2/x^4 - 3 y'/x

Substituting the value of y' from equation (2),

y ' ' = -2/x^4 - 3 * (1 - 2 ln(4x))/x^4

y ' ' = -2/x^4 - 3/x^4 + 6 ln(4x)/x^4

y ' ' = -5/x^4 + 6ln(x)/x^4

y ' ' = (6 ln(x) - 5)/x^4

Ans: y' = (1 - 2 ln(4x))/x^3

y ' ' = (6 ln(x) - 5)/x^4

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akhl

@akhl

Dazzler

Posted: 15 years ago

#162

Use logarithmic differentiation to find the derivative of the function.

y=[(sin(x))^2 (tan(x))^6]/(x^2+3)^2

ln y = 2 ln sin(x) + 6 ln tan(x) - 2 ln(x^2 + 3)

Differentiating both sides with respect to x,

y'/y = 2 * cos(x)/sin(x) + 6 * sec^2(x)/tan(x) - 2 * 2x/(x^2 + 3)

y'/y = 2/tan(x) + 6 sec^2(x)/tan(x) - 4x/(x^2 + 3)

y' = y[2/tan(x) + 6 sec^2(x)/tan(x) - 4x/(x^2 + 3)]

y' = [(sin(x))^2 (tan(x))^6]/(x^2+3)^2 * [2/tan(x) + 6 sec^2(x)/tan(x) - 4x/(x^2 + 3)]

y' = 2 sin^2(x) tan^5(x)/(x^2+3)^2 + 6 sin^2(x) tan^6(x) sec^2(x)/tan(x)/(x^2+3)^2 - 4x sin^2(x) tan^6(x)/(x^2+3)^3

y' = 2 sin^2(x) tan^5(x)/(x^2+3)^2 +6 sin^2(x) sec^2(x) tan^5(x)/(x^2+3)^2 - 4x sin^2(x) tan^6(x)/(x^2+3)^3

y' = 2 sin^2(x) tan^5(x)/(x^2+3)^2 + 6 sin^2(x) 1/cos^2(x) tan^5(x)/(x^2+3)^2 - 4x sin^2(x) tan^6(x)/(x^2+3)^3

y' = 2 sin^2(x) tan^5(x)/(x^2+3)^2 + 6 tan^2(x) tan^5(x)/(x^2+3)^2 - 4x sin^2(x) tan^6(x)/(x^2+3)^3

y' = 2 sin^2(x) tan^5(x)/(x^2+3)^2 + 6 tan^7(x)/(x^2+3)^2 - 4x sin^2(x) tan^6(x)/(x^2+3)^3

y' = {[2 *sin^2(x) tan^5(x) + 6 tan^7(x)](x^2 + 3) - 4(x)sin^2(x)tan^6(x)}/(x^2 + 3)^3

Ans: y' = {[2 *sin^2(x) tan^5(x) + 6 tan^7(x)](x^2 + 3) - 4(x)sin^2(x)tan^6(x)}/(x^2 + 3)^3

Or if it is ok to write answer in terms of both x and y, then

y' = y[2/tan(x) + 6 sec^2(x)/tan(x) - 4x/(x^2 + 3)]

Edited by akhl - 15 years ago

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Sonia

@Miggi

Inactive

Posted: 15 years ago

#163

Thanks a lot for ur help!😊🤗👏 could u plz explain to me how did u get this part of last ques?

y' = {[2 *sin^2(x) tan^5(x) + 6 tan^7(x)](x^2 + 3)^2 - 4xsin^2(x)tan^6(x)}/(x^2 + 3)^4

I would really appreciate that.

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akhl

@akhl

Dazzler

Posted: 15 years ago

#164

See my previous answer. I mean my answer before your comment. I have edited it to show all the steps.

There was also some mistake with exponent. Corrected it.

Edited by akhl - 15 years ago

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Sonia

@Miggi

Inactive

Posted: 15 years ago

#165

Originally posted by: akhl
y' = {[2 *sin^2(x) tan^5(x) + 6 tan^7(x)](x^2 + 3)^2 - 4xsin^2(x)tan^6(x)}/(x^2 + 3)^4

Oh I skipped some steps.

[(sin(x))^2 (tan(x))^6]/(x^2+3)^2 * 2/tan(x) is

2 sin^2(x) tan^5(x)/(x^2+3)^2

[(sin(x))^2 (tan(x))^6]/(x^2+3)^2 * 6 sec^2(x)/tan(x) is

6 sin^2(x) sec^2(x) tan^5(x)/(x^2+3)^2

= 6 sin^2(x)/cos^2(x) * tan^5(x)/(x^2+3)^2

= 6 tan^2(x) * tan^5(x)/(x^2+3)^2

= 6 tan^5(x)/(x^2+3)^2

[(sin(x))^2 (tan(x))^6]/(x^2+3)^2 * 4x/(x^2+3)^2 is

4x[(sin(x))^2 (tan(x))^6]/(x^2+3)^3

Then I took (x^2+3)^3 as LCM

Ohh, make one correction. I wrote (x^2+3)^4

Change it to (x^2+3)^3

Other things are correct.

Yea okay thnxx thts what i was wondering, but for question y= (ln(x))^(cos(15x)) the answer u gave me was the same as i got but its saying tht its not correct. plus in this question when i post the answer the site says check ur syntax i've tried different thingd but its still saying the same thing :/

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akhl

@akhl

Dazzler

Posted: 15 years ago

#166

Correct answer to the last question is

y' = {[2 *sin^2(x) tan^5(x) + 6 tan^7(x)](x^2 + 3) - 4(x)sin^2(x)tan^6(x)}/(x^2 + 3)^3

Or if it is ok to write answer in terms of both x and y, then

y' = y[2/tan(x) + 6 sec^2(x)/tan(x) - 4x/(x^2 + 3)]

I edited my original answer to correct it and show all the steps.

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Sonia

@Miggi

Inactive

Posted: 15 years ago

#167

nd also isn't sinx/cosxx =tan and cosx/sinx =cotx i am really confused its in step 2 of last ques

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Sonia

@Miggi

Inactive

Posted: 15 years ago

#168

ohh i got it we jus have to write 2cotx in instead of tanx in last step.

could u plzz check the other ques too cuz the site is saying its wrong

y= (ln(x))^(cos(15x))

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akhl

@akhl

Dazzler

Posted: 15 years ago

#169

^{I will try differently this time y = (ln(x))^cos(15x)
ln y = cos(15x) ln[ln(x)]
Differentiating with respect to x,
y'/y = cos(15x) * 1/ln(x) * 1/x - 15 sin(15x) ln(ln(x))
y'/y = cos(15x)/(x ln(x)) - 15 sin(15x) ln(ln(x))
y'/y = (cos(15x)- 15 (x) sin(15x) ln(x) ln(ln(x)))/(x ln(x))
y = y (cos(15x)- 15 (x) sin(15x) ln(x) ln(ln(x)))/(x ln(x))
y' = (ln(x))^cos(15x) * (cos(15x)- 15 (x) sin(15x) ln(x) ln(ln(x)))/(x ln(x))

dont know if it can be simplified further,\.
I have used only ( ). See if website expects [ ] or { }}