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Question #c9699 - Socratic Explanation: #e^y=lnx+sin^3 (sqrt (3x-5))# We can differentiate both sides: #e^ydy= (1 x+3sin^2 (sqrt (3x-5))cos (sqrt (3x-5))1 (2sqrt (3x-5)) (3))dx# Now, we can
Notation for the Second Derivative - Socratic How do you find the first and second derivative of # (lnx) x^2#? How do you find the first and second derivative of #lnx^ (1 2)#? How do you find the first and second derivative of #x (lnx)^2#? How do you find the first and second derivative of #ln (x^2-4)#? How do you find the first and second derivative of #ln (lnx^2)#?
Is the function concave up or down if #f (x)= (lnx)^2#? - Socratic It is concave up on the interval (0,e) and concave down on (e, oo) f (x)= (lnx)^2 f' (x) = 2 (lnx)*1 x = (2lnx) x f'' (x) = ( (2 x)*x - 2lnx *1) x^2 = (2 (1-lnx)) x^2 1-lnx = 0 where lnx = 1 which is at x=e All other factors of f'' (x) are always positive, so the sign of f'' (x) is the same as the sign of 1-lnx That is: f'' (x) is positive if x < e (so lnx < 1) and it is negative for x
How do I determine if #int_1^oolnx x^2 dx# converges or diverges? The integral is convergent and converges to 1 To calculate this improper integral, first calculate I=int_1^a (lnxdx) x^2 Perform this integral by parts u'=1 x^2, =>, u=-1 x v=lnx, =>, v'=1 x intu'v=uv-intuv' Therefore, I=int_1^a (lnxdx) x^2= [-lnx x]_1^a+int_1^adx x^2 = [-lnx x]_1^a- [1 x]_1^ a = (-lna a-0)- (1 a-1) =-lna a-1 a+1 Now, calculate the limits lim_ (a->oo) (-lna a-1 a+1)=0-0+1=1