Using logarithmic differentiate technique, find dy/dx
y = [(x-3)³(x²+1) / (2x+5)³] raised to the power of 1/5
I don't know how to type this. It is the 5th root of all those.
You have
y = [(x-3)³(x²+1) / (2x+5)³]1/5
so
ln y = (1/5)ln([(x-3)³(x²+1) / (2x+5)³] =
(1/5){ln([(x-3)³(x²+1)) - ln(2x+5)³}=
(1/5){3ln(x-3) + ln(x²+1) - 3ln(2x+5)}
Now find dy/dx and exponentiate your result to get the final result.
I hope I read your superscripts right, they're a little small for me.
To find dy/dx using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm of both sides of the equation.
ln y = (1/5){3ln(x-3) + ln(x²+1) - 3ln(2x+5)}
Step 2: Use the properties of logarithms to simplify the expression.
ln y = (1/5){ln[(x-3)³(x²+1)] - ln(2x+5)³}
Step 3: Apply the power rule of logarithms to expand the expression inside the parentheses.
ln y = (1/5){ln[(x-3)³] + ln(x²+1) - ln[(2x+5)³]}
Step 4: Apply the power rule and product rule to simplify further.
ln y = (1/5){3ln(x-3) + ln(x²+1) - 3[ln(2x+5)]}
Step 5: Differentiate both sides of the equation with respect to x.
1/y * (dy/dx) = (1/5){(3/(x-3)) + (2x/(x²+1)) - 3(2/(2x+5))}
Step 6: Multiply both sides by y to isolate (dy/dx).
dy/dx = y * (1/5){(3/(x-3)) + (2x/(x²+1)) - 3(2/(2x+5))}
Step 7: Substitute y back into the expression.
dy/dx = [(x-3)³(x²+1) / (2x+5)³]^(1/5) * (1/5){(3/(x-3)) + (2x/(x²+1)) - 3(2/(2x+5))}
That is the expression for dy/dx using the logarithmic differentiation technique.