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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.
What is equivalent to ln 6 plus ln 4?
You can also write this as ln(6 times 4)
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 2 divided by x as a power of x?
x^(ln(2)/ln(x)-1)
What is the derivative of lnlnx?
1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]
What is the derivative of xlnx?
ln(x)+1. This can be found using the product rule: if f(x)=uv, f'(x)=u'v+v'u, so if u is x and v is ln(x), the product rule gives (1)(ln(x))+(x)(1/x)=ln(x)+1.
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.
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 derivitive of ln10x?
The derivative of ln(x) is 1/x. Therefore, by Chain Rule, we get:[ln(10x)]' = 1/10x * 10 = 1/xUsing this method, you can also infer that the derivative of ln(Ax) where A is any constant equals 1/x.
What is the derivative of ln x to the power of 2?
If the function is (ln x)2, then the chain rules gives us the derivative 2ln(x)/x, with the x in the denominator. If the function is ln (x2), then the chain rule gives us the derivative 2/x.
How do you Differentiate and Integrate y equals 3 to the power of x?
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
What is the derivative of Ln 2 plus x?
The order of operations is not quite clear here.If you mean (ln 2) + x, the derivate is 0 + 1 = 1.If you mean ln(2+x), by the chain rule, you get (1/x) times (0+1) = 1/x.
What is the derivative of half to the power of x?
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
What is the derivative of a half to the power of x?
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
How would you solve ln 4 plus 3 ln x equals 5 ln 2?
Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2
What is the derivative of ln 1-x?
In this case, you need to apply the chain rule. Note that the derivative of ln N = 1/N. In that case we get: f(x) = ln(1 - x) ∴ f'(x) = 1/(1 - x) × -1 ∴ f'(x) = -1/(1 - x)
