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By using the basic rules of exponents, plus the fact that the exponential function (e raised to some power) and the natural logarithm are inverse functions.
e8 ln x + cos x
= e8 ln x ecosx
= e(ln x)(8) ecosx
= (eln x)8 ecosx
= (eln x)8 ecosx
= x8 ecosx
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Integral of sin 2x divided by 1 plus cosine squared x?
3
What is equivalent to ln 6 plus ln 4?
You can also write this as ln(6 times 4)
What is the derivative of x to the power of ln x?
I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.
Integrate of ln x squared?
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
How do you differentiate exponential function?
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
What is ten raised to the with power?
Using Euler's relation, we know that e^(i*n*pi) = cos(n*pi) + i*sin(n*pi) where n is an integer. We also know that we can rewrite 10 as e raised to a specific power, namely e^(ln(10)). So substituting this back into 10^i and then applying Euler's relation, we obtain 10^i = (e^(ln(10)))^i = (e^(i*ln(10))) = cos(ln(10)) + i*sin(ln(10)).
Simplify e raised to the power of lnx plus lny?
There are 2 interpretations of your question: First: e^[lnx + lny] =e^[ln(xy)] =xy Second: lny + e^(lnx) =lny + x
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
What is the integration of tanx?
The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
What is the derivative of cos lnx?
This is a chain rule question. Let u = ln(x) d{cos[ln(x)]}/dx = (d[cos(u)]/du)*(du/dx) = -sin(u)*(du/dx) = -sin[ln(x)]*d[ln(x)]/dx = -sin[ln(x)]/x
What is the anti-derivative of a tangent function?
-ln|cos x| + C
Simplify In e to the 7th power?
It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.
How do you solve lntan(x)lnsin(x)-lncos(x)?
How do you solve ln|tan(x)|=ln|sin(x)|-ln|cos(x)|? Well you start by........
What is the derivative of sinx pwr cosx?
For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.
What is integral of sin x by 1 plus cos x?
Int( sin(x) / ( 1+cos(x) ) ) = Int (-du/u) Perform a u-substitution, u=1+cosx du=d/dx(1+cosx)=-sinx =-ln |u| + C negative pulls out of integral (Don't forget the constant) =ln|1+cosx|+C
Integral of sin 2x divided by 1 plus cosine squared x?
3
