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y = ln(tan(x)) u = tanx y =ln(u) dy/du = 1/u du/dx = sec2(x) dy/dx = dy/du * du/dx = sec2(x)/tan(x)

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โˆ™ 2009-03-18 11:54:50
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Q: Differential co-efficient of log tan x is?
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How do you prove that the derivative of sec x is equal to sec x tan x?

Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.


What is 525677.42742534 to the 85 power?

Since this goes beyond the capability of most calculators, and of Excel too, you will have to get an approximation with logarithms. Better use base 10 logarithms; in Excel you can use the log() function. Call your number "x":x = 525677.4274253485log x = log 525677.4274253485Now use a standard property of logarithms:log x = 85 log 525677.42742534log x = (85)(5.72071932882537)log x = 486.261142950157So, the result is approximately 10486. To get the coefficient more accurately:x = 10486.261142950157x = 10486 + 0.261142950157x = 100.26114295015710486x = 1.82449614534211 x 10486Since this goes beyond the capability of most calculators, and of Excel too, you will have to get an approximation with logarithms. Better use base 10 logarithms; in Excel you can use the log() function. Call your number "x":x = 525677.4274253485log x = log 525677.4274253485Now use a standard property of logarithms:log x = 85 log 525677.42742534log x = (85)(5.72071932882537)log x = 486.261142950157So, the result is approximately 10486. To get the coefficient more accurately:x = 10486.261142950157x = 10486 + 0.261142950157x = 100.26114295015710486x = 1.82449614534211 x 10486Since this goes beyond the capability of most calculators, and of Excel too, you will have to get an approximation with logarithms. Better use base 10 logarithms; in Excel you can use the log() function. Call your number "x":x = 525677.4274253485log x = log 525677.4274253485Now use a standard property of logarithms:log x = 85 log 525677.42742534log x = (85)(5.72071932882537)log x = 486.261142950157So, the result is approximately 10486. To get the coefficient more accurately:x = 10486.261142950157x = 10486 + 0.261142950157x = 100.26114295015710486x = 1.82449614534211 x 10486Since this goes beyond the capability of most calculators, and of Excel too, you will have to get an approximation with logarithms. Better use base 10 logarithms; in Excel you can use the log() function. Call your number "x":x = 525677.4274253485log x = log 525677.4274253485Now use a standard property of logarithms:log x = 85 log 525677.42742534log x = (85)(5.72071932882537)log x = 486.261142950157So, the result is approximately 10486. To get the coefficient more accurately:x = 10486.261142950157x = 10486 + 0.261142950157x = 100.26114295015710486x = 1.82449614534211 x 10486


Sec x times sin x divided by tan x?

1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1


What is the derivative of -x-y?

If y is a function of x, that is y=f(x), then the derivative of x-y is 1-y' or 1-dy/dx (where y' or dy/dx is the differential coefficient of y with respect to x).


What is the answer to cot squared x - tan squared x equals 0?

cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.


Integral of cos x cot x?

cos(x) cot(x) = cos(x) * 1/(tan(x)) = cos(x) * 1 / (sinx(x) / cos(x)) = cos2(x) / sin(x) = (1-sin2(x)) / sin(x) = 1/sin(x) - sin(x) so the antiderivative of cos(x)cot(x) = log[abs(tan(x/2))]+cos(x) This can also be written as log[abs((sin(x)/(cos(x)+1))]+cos(x) if we want everything in terms of x and not (x/2). The two answers are, of course, the same. where log(x) refers to the natural log, often written ln(x). We might write ln[|sin(x)|/|cos(x)+1|] +cos(x)


What is the answer to log x6?

log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)


Which expression has the same value as tan(-x) for all values for x?

tan(-x) = -tan(x)


Sin x Tan x equals Sin x?

No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)


How can arccot of tanx be simplified?

There is not much that can be done by way of simplification. Suppose arccot(y) = tan(x) then y = cot[tan(x)] = 1/tan(tan(x)) Now cot is NOT the inverse of tan, but its reciprocal. So the expression in the first of above equation cannot be simplified further. Similarly tan[tan(x)] is NOT tan(x)*tan(x) = tan2(x)


What is the coefficient of x in 4 plus 3y plus x?

The coefficient of X is 1


What is tan x csc x?

tan(x)*csc(x) = sec(x)


What is cos x tan x simlpified?

The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).


Log x plus log 2 equals log 2?

log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1


What is log to the third power equal to?

Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.


What is a constant coefficient?

In a ploynomial or differential equation or really any formula or equation with variables in it, the coefficients are the terms "in front of" the variable or multiplied the variables. Each variable generally has its own coefficient. If a coefficient is constant (ie just a number) then it is a constant coefficient. eg Consider the polynomial , 3x2+9yx+6 in terms of x. It has one constant coefficient (3), one variable coefficient (9y) and one constant (6).


Log x - log 6 equals log 15?

log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90


Log9x plus log x equals 4log10?

log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3


What is the period of y equals tan2x?

The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2


What is the derivative of ln cos x?

The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).


How do you solve log x plus 2 equals log 9?

log x + 2 = log 9 log x - log 9 = -2 log (x/9) = -2 x/9 = 10^(-2) x/9 = 1/10^2 x/9 = 1/100 x= 9/100 x=.09


What are 3 logarithmic properties?

log(1) = 0log(x*y) = log(x) + log(y)If logb(x) = y then x = by.


If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?

tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).


How do you prove tan x plus tan x sec 2x equals tan 2x?

tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x


Rules of log?

Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0