What is the second derivative of y=e^5x+(lnx)/2x ?
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We have to find the second derivative of y = e^5x + (ln x) / 2x
y = e^5x + (ln x) / 2x
=> y = e^5x + (1/2)(ln x)*(1/x)
y' =...
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To determine the second derivative, we'll have to determine the 1st derivative, for the beginning. We'll differentiate y with respect to x.
dy/dx = d(e^5x+(lnx)/2x)/dx
We'll apply chain rule for the first term of the sum and the quotient rule for the 2nd terms of the sum.
dy/dx = 5e^5x + [(lnx)'*2x - (lnx)*(2x)']/4x^2
dy/dx = 5e^5x + (2x/x - 2lnx)/4x^2
dy/dx = 5e^5x + 2(1 - lnx)/4x^2
dy/dx = 5e^5x + (1 - lnx)/2x^2
dy/dx = 5e^5x + (lne - lnx)/2x^2
dy/dx = 5e^5x + [ln(e/x)]/2x^2
Now, we'll determine the second derivative:
d^2y/dx^2 = [d[5e^5x + [ln(e/x)]/2x^2/dx
]In other words, we'll determine the derivative of the expression of the 1st derivative:
d^2y/dx^2= 25e^5x + {-2x^2/x - [1-ln(x]*4x}/4x^4
d^2y/dx^2 = 25e^5x - 2x(1+2-2lnx)/4x^4
d^2y/dx^2 = 25e^5x - 2x(1+2-2lnx)/4x^4
d^2y/dx^2 = 25e^5x - (1+2-2lnx)/2x^3
d^2y/dx^2 = 25e^5x - (3 - lnx^2)/2x^3
The second derivative is: d^2y/dx^2 = 25e^5x - (3 - lnx^2)/2x^3.
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