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How to calculate the n-th power of 1.1 equal to 2
(1) Calculation method: 1.1 ^ n = 2 take common logarithm on both sides: n log1.1 = log2n = log2 / log1.1 ≈ 0.3010299956639521373889472449 / 0.041392685158225040750199971243024 = 7.272540899734171908331990367496 in practical application, it can be according to specific requirements
Find the derivative of y = x / 1-cosx
y'=[x'(1-cosx)-x(1-cosx)']/(1-cosx) ²
=(1-cosx-xsinx)/(1-cosx) ²
Find the derivative of y = cosx / X It's a tragedy... Why do I always calculate (- SiNx * x + cosx) / 2x
The derivative division rule is (U / V) '= (u'v-v'u) / (V ^ 2). You're wrong. First, there's a minus sign in the middle; Secondly, the denominator is square, and the correct form should be y '= (cosx / x)' = (- SiNx * x - cosx * 1) / X ² = - (cosx + xsinx) / x ² Or use the derivative multiplication rule (UV) '= u
Y = (cosx) ^ x derivative
y=(cosx)^x
lny=xln(cosx)
Simultaneous derivation on both sides
y'/y=ln(cosx)+x*(-sinx)/cosx
y'=(cosx)^x*[ln(cosx)-x*tanx]
Find the derivative of y = cosx / SiNx
Just follow the formula
y'=[(cosx)'sinx-cosx(sinx)']/(sinx)^2
=[-sinxsinx-cosxcosx]/(sinx)^2
=-1/(sinx)^2
e^1/3*(lna+lnb+lnc)=(abc)^1/3? How can I turn it into (ABC) ^ 1 / 3,
lna+lnb+lnc=ln(abc)
Original formula = e ^ [1 / 3ln (ABC)] = e ^ ln [(ABC) ^ 1 / 3] = (ABC) ^ 1 / 3
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