Use logarithmic differentiation to find dy/dx.
y = sq. root [(x^2 - 1)/(x^2 + 1)]
To find the derivative dy/dx using logarithmic differentiation, we will follow these steps:
Step 1: Take the natural logarithm of both sides of the equation:
ln(y) = ln[(x^2 - 1)/(x^2 + 1)]
Step 2: Use the properties of logarithms to simplify the expression:
ln(y) = ln(x^2 - 1) - ln(x^2 + 1)
Step 3: Differentiate both sides of the equation implicitly with respect to x:
(1/y) * dy/dx = (1/(x^2 - 1)) * 2x - (1/(x^2 + 1)) * 2x
Step 4: Simplify the right-hand side of the equation:
(1/y) * dy/dx = (2x/(x^2 - 1) - 2x/(x^2 + 1))
Step 5: Multiply both sides of the equation by y:
dy/dx = y * (2x/(x^2 - 1) - 2x/(x^2 + 1))
Step 6: Substitute the value of y back into the equation:
dy/dx = sq. root [(x^2 - 1)/(x^2 + 1)] * (2x/(x^2 - 1) - 2x/(x^2 + 1))
Therefore, the derivative dy/dx of y = sq. root [(x^2 - 1)/(x^2 + 1)] using logarithmic differentiation is dy/dx = sq. root [(x^2 - 1)/(x^2 + 1)] * (2x/(x^2 - 1) - 2x/(x^2 + 1)).
To find the derivative of y = sqrt[(x^2 - 1)/(x^2 + 1)] using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm of both sides of the equation:
ln(y) = ln[sqrt((x^2 - 1)/(x^2 + 1))]
Step 2: Simplify the logarithm using the properties of logarithms:
ln(y) = (1/2) * ln[(x^2 - 1)/(x^2 + 1)]
Step 3: Differentiate both sides of the equation implicitly with respect to x:
(d/dx)[ln(y)] = (d/dx)[(1/2) * ln[(x^2 - 1)/(x^2 + 1)]]
Step 4: Apply the chain rule on the left side:
(dy/dx) * (1/y) = (1/2) * (d/dx)[ln[(x^2 - 1)/(x^2 + 1)]]
Step 5: Find the derivatives on the right side:
(dy/dx) * (1/y) = (1/2) * [(1/(x^2 - 1)/(x^2 + 1))] * [(d/dx)[(x^2 - 1)/(x^2 + 1)]]
Step 6: Simplify the right side expression using the quotient rule:
(dy/dx) * (1/y) = (1/2) * [(x^2 + 1)(2x) - (x^2 - 1)(2x)] / [(x^2 + 1)^2]
Step 7: Simplify the expression further:
(dy/dx) * (1/y) = (2x^3 + 2x) / [(x^2 + 1)^2]
Step 8: Multiply both sides of the equation by y:
dy/dx = y * (2x^3 + 2x) / [(x^2 + 1)^2]
Step 9: Substitute the value of y:
dy/dx = sqrt[(x^2 - 1)/(x^2 + 1)] * (2x^3 + 2x) / [(x^2 + 1)^2]
Thus, the derivative dy/dx of y = sqrt[(x^2 - 1)/(x^2 + 1)] is given by:
dy/dx = sqrt[(x^2 - 1)/(x^2 + 1)] * (2x^3 + 2x) / [(x^2 + 1)^2]
y = sq. root [(x^2 - 1)/(x^2 + 1)]
take log (to base e, =ln) on both sides
log(y) = (1/2)(log(x²-1)-log(x²+1)
differentiate with respect to x
dy/dx / y = x/(x²-1)-x/(x²+1)
= 2x/((x²-1)(x²+1))
dy/dx = x/(x²-1)-x/(x²+1)
= 2x/((x²-1)(x²+1)) * √(x²-1)/√(x²+1)
Simplify.