# Differentiate the following with respect to x. ln(sinxcosx) Thanks!

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

we can further simplify this:

cotx - tanx = `cosx/sinx - sinx/cosx = (cos^2x-sin^2x)/(sinxcosx)`

multiplying the numerator and denominator by 2, we get

`2(cos^2x - sin^2x)/(2sinxcosx) = (2cos2x)/(sin2x) = 2 cot2x`

we have used the identities: cos2x = cos^2x- sin^2x and sin2x = 2sinxcosx.

hope this helps.

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

we can use the logarithmic property: ln (ab) = ln a + ln b

thus, `d/dx ln(sinx cosx) = d/dx ln sinx + d/dx ln cosx`

`= 1/sinx d/dx sinx + 1/cosx d/dx cosx = cosx/sinx - sinx/cosx = cot x - tan x`

Hope this helps.

phanpal999 | Student, Grade 8 | (Level 1) Salutatorian

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kseddy123 | College Teacher | (Level 1) Assistant Educator

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ln(sinx . cos x)

Multiply and divide sinx . cos x by 2

= ln(`(2.sinx .cosx)/(2)` ` ` )

= ln( `sin(2x)/2` )

= [ln( ` ` sin(2x)) -  ln(2)]

=[ln( sin(2x))] - [ ln(2)]

differentiation of a constant is 0 is [ ln(2)] =0

=[ln( sin(2x))]

=` ``1/sin(2x) ` sin(2x) = `2cos(2x)/sin(2x) ` = 2 cot(2x)   <----- is the answer

kseddy123 | College Teacher | (Level 1) Assistant Educator

Posted on

i gave d/dx .....but its missing every where so .....i am re writing this again

`d/dx` ln(sinx . cos x)

Multiply and divide sinx . cos x by 2

= `d/dx ` ln( )

= `d/dx `ln( )

= `d/dx `[ln( sin(2x)) -  ln(2)]

=`d/dx `[ln( sin(2x))] -`d/dx ` [ ln(2)]

differentiation of a constant is 0 is [ ln(2)] =0

=`d/dx `[ln( sin(2x))]

= `d/dx `sin(2x) = = 2 cot(2x)   <----- is the answer

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kspcr111 | In Training Educator

Posted on

Hi

till `d/dx` ln(sinx . cos x)= (cot x- tanx )

the above steps are the same .......plz check the attachment for further steps to get the answer which u required

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