Find the second derivative of the following function:
y=log(base2)x/x²
y'=[(x/ln2)-2xlog(base2)x]/x^4
the derivative of x/ln2 is (ln2)^-1?
y"={[((ln2)^(-1))-(2log(base2)x/xln2)]x^4-4x³[(x/ln2)-2xlog(base2)x]}/x^8
I tried simplifying it and I got,
y"=[3/(x^4)ln2]-[2long(base2)x/(x^5)ln2]+[8log(base2)x/x^4]
Is the answer really off?
Thanks.
what a messy quesstion!
In your y'' , your forgot to treat the 2xlog2x as a product rule
I had
y' = [ x^4(1/ln2 - (2x(1/xln2) + 2log2x) - 4x^3(x/ln2 - 2xlog2x ]/x^8
= x^3[ x/ln2 - 2x/ln2 + 2xlog2x - 4/ln2 + 8xlog2x ]/x^8
= (10xlog2x - 5x/ln2)/x^5
check my steps to see if I am close.
Shouldn't it be -2xlog(base2)x because the product rule is inside the bracket, and its subtracting the bracket.
y"=[x/ln - 2xlog(base2)x -2x²/xln2 - 4x/ln2 + 8xlog(base2)x]/x^5
=[-3x/x^5] + [6xlog(base2)x/x^5] - [2x²/xln2]
Then I put them in common denominator,
[-5+6(log(base2)x)(ln2)]/(x^4)(ln2).
Thanks for guiding me!
To find the second derivative of the function y = log(base2)x/x², you correctly found the first derivative as y' = [(x/ln2) - 2xlog(base2)x] / x^4. However, there seems to be some confusion in finding the second derivative.
To clarify, let's go through the steps to find the second derivative:
Step 1: Differentiate y' with respect to x.
To differentiate y' with respect to x, we can use the quotient rule. The quotient rule states that for functions u(x) and v(x), the derivative of u(x)/v(x) is (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2.
In this case, u(x) = (x/ln2) - 2xlog(base2)x and v(x) = x^4.
Now, let's differentiate u(x) and v(x) separately:
u'(x) = d/dx[(x/ln2) - 2xlog(base2)x]
= (1/ln2) - 2[1 + log(base2)x]
= 1/ln2 - 2 - 2log(base2)x
v'(x) = d/dx[x^4]
= 4x^3
Using the quotient rule, the second derivative y" becomes:
y" = [(v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2] / x^4
= [(x^4 * (1/ln2 - 2 - 2log(base2)x) - ((x/ln2) - 2xlog(base2)x) * 4x^3) / (x^4)^2] / x^4
Now, we can simplify this expression:
y" = [(x^4/ln2 - 2x^5 - 2x^4log(base2)x) - (4x^4/ln2 - 8x^4log(base2)x)] / x^8
= [(x^4/ln2 - 4x^4/ln2) - (2x^5 - 8x^4log(base2)x)] / x^8
= [(-3x^4/ln2) - (2x^5 - 8x^4log(base2)x)] / x^8
= (-3x^4 - 2x^5 + 8x^4log(base2)x) / (ln2 * x^8)
So, the second derivative of y = log(base2)x/x² is given by y" = (-3x^4 - 2x^5 + 8x^4log(base2)x) / (ln2 * x^8).
Therefore, your attempted simplification was not correct. The correct expression for the second derivative is y" = (-3x^4 - 2x^5 + 8x^4log(base2)x) / (ln2 * x^8).