In (x^(ln2)), ln2 is a constant, so the Power Rule can be used. d/dx (x^ln2)=(ln2)x^(ln2-1)
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
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The half-life of a radioactive nuclide when 95% of it is left after one year is 13.5 years. AT = A0 2(-T/H) 0.95 = (1) 2(-1/H) ln2(0.95) = -1/H H = -1/ln2(0.95) H = 13.5
In (x^(ln2)), ln2 is a constant, so the Power Rule can be used. d/dx (x^ln2)=(ln2)x^(ln2-1)
ln2^x = xln2. let ln2 = k (constant), then the differential = k. Hence d(ln2^x)/dx = ln2
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
LN2 i think :D
Very much
The integral of ln(2) is a constant multiple of x times the natural logarithm of 2, plus a constant of integration. In other words, the integral of ln(2) with respect to x is x * ln(2) + C, where C is the constant of integration. This integral represents the area under the curve of the natural logarithm of 2 function with respect to x.
LN2 other know as liquid nitrogen is a solution that can make water hard
3lnx - ln2=4 lnx^3 - ln2=4 ln(x^3/2)=4 (x^3)/2=e^4 x^3=2e^4 x=[2e^4]^(1/3)
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The half-life of a radioactive nuclide when 95% of it is left after one year is 13.5 years. AT = A0 2(-T/H) 0.95 = (1) 2(-1/H) ln2(0.95) = -1/H H = -1/ln2(0.95) H = 13.5
ln is the inverse of e. So the e and the ln cancel each other out and you are left with 2. eln2 = 2