How do you differentiate # y =-ln [ 3+ (9+x^2) / x]#?

Question

How do you differentiate # y =-ln [ 3+ (9+x^2)  / x]#?

Ezra Adams · Accepted Answer

#dy/dx=-1/(3+9/x+x)*(1-9/x^2)# #y=-ln(3+(9+x^2)/x)=-ln(3+9/x+x)# Use the chain rule #dy/dx=dy/(du)*(du)/dx# Let #y=-ln(u)# then #dy/(du)=-1/u# And #u=3+9/x+x# then #(du)/(dx)=-9/x^2+1# #dy/dx=-1/u*(1-9/x^2)# Substitute #u=3+9/x+x# #dy/dx=-1/(3+9/x+x)*(1-9/x^2)#

Adam Adkins · Answer

undefined To differentiate the given function y = -ln[3 + (9 + x^2) / x], you can use the chain rule and the properties of natural logarithms:

1. Let u = 3 + (9 + x^2) / x.
2. Then y = -ln(u).
3. Find the derivative of u with respect to x: du/dx.
4. Apply the chain rule: dy/dx = dy/du * du/dx.
5. Find dy/du: -1/u.
6. Combine steps 4 and 5: dy/dx = (-1/u) * du/dx.

Now, find du/dx by differentiating u with respect to x.

7. Use the quotient rule for differentiation.

$$ \frac{d}{dx} \left( \frac{9 + x^2}{x} \right) = \frac{(2x)(x) - (9 + x^2)(1)}{x^2} = \frac{2x^2 - 9 - x^2}{x^2} = \frac{x^2 - 9}{x^2} $$

8. Substitute du/dx and u back into the expression for dy/dx.

$$ dy/dx = \frac{-1}{3 + \frac{9 + x^2}{x}} \times \frac{x^2 - 9}{x^2} $$

$$ dy/dx = \frac{-(x^2 - 9)}{x(x + \frac{9 + x^2}{x})} $$

$$ dy/dx = \frac{-(x^2 - 9)}{x^2 + 9 + x} $$