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Let y = ln(x2), where ln is log to base e.
Using our log laws, y = 2 ln (x)
Therefore dy/dx = 2 . 1/x = 2/x
Another way to do this is to use the chain rule (for differentiation)
Let y = ln(x2), where ln is log to base e.
Let u = x2, therefore du/dx = 2x
y = ln u, therefore dy/du = 1/u
Putting it altogether gives:
dy/dx = dy/du . du/dx = 1/u . 2x = 2x/x2 = 2/x
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Is logarithmic differentiation specially useful when dealing with functions with exponent that also depend on x?
Yes. For example, to differentiate y = (x^2 + 1)^x, we take the natural log of both sides.ln(y) = ln((x^2 + 1)^x) Bring down the exponent. ln(y) = x ln(x^2 + 1) Differentiate both sides. dy/y = ((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Substitute in y = (x^2 + 1)^x. dy/((x^2 + 1)^x) =((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Solve for dy/dx. dy/dx = ((x^2 + 1)^x)((2x^2)/(x^2 + 1) + ln(x^2 + 1))
What is the antiderivative of e to the power of one divided by x?
1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace
What is the anti-derivative of 5 to the x power?
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
2 to the power of x equals 5 what is x?
To find the value of x when 2^x = 5, we can take the logarithm of both sides. Using the natural logarithm (ln) gives us: ln(2^x) = ln(5). Using the property of logarithms that allows us to bring the exponent down as a multiplier, we get: x*ln(2) = ln(5). Finally, dividing both sides by ln(2) gives us the value of x: x = ln(5)/ln(2), which is approximately 2.322.
How do you differentiate 2 to the power x with respect to x?
The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.
Is logarithmic differentiation specially useful when dealing with functions with exponent that also depend on x?
Yes. For example, to differentiate y = (x^2 + 1)^x, we take the natural log of both sides.ln(y) = ln((x^2 + 1)^x) Bring down the exponent. ln(y) = x ln(x^2 + 1) Differentiate both sides. dy/y = ((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Substitute in y = (x^2 + 1)^x. dy/((x^2 + 1)^x) =((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Solve for dy/dx. dy/dx = ((x^2 + 1)^x)((2x^2)/(x^2 + 1) + ln(x^2 + 1))
What is the antiderivative of e to the power of one divided by x?
1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace
What is the anti-derivative of 5 to the x power?
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
What is the derivative of lnx2?
Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x
Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
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.
How do you differentiate ln 2 to the exponent x?
ln2^x = xln2. let ln2 = k (constant), then the differential = k. Hence d(ln2^x)/dx = ln2
2 to the power of x equals 5 what is x?
To find the value of x when 2^x = 5, we can take the logarithm of both sides. Using the natural logarithm (ln) gives us: ln(2^x) = ln(5). Using the property of logarithms that allows us to bring the exponent down as a multiplier, we get: x*ln(2) = ln(5). Finally, dividing both sides by ln(2) gives us the value of x: x = ln(5)/ln(2), which is approximately 2.322.
What is 2 divided by x as a power of x?
x^(ln(2)/ln(x)-1)
Differentiate y equals a power x?
y=ax y'=ln(a)*ax
How do you solve log2 equals x?
log(2) = X can be expressed exponentially like this, because by the definition of logs( base 10) this is what this means. 10^X = 2 take natural log each side ln(10^X) = ln(2) you have right to place X in front of ln X ln(10) = ln(2) X = ln(2)/ln(10) ( not ln(2/10)!! ) X = 0.3010299957 check 10^0.3010299957 = 2 checks
