6. If y = ln((1+x)/(1-x)), find dy/dx and simplify
Using the chain rule with ln(u) where u = (1+x)/(1-x):
dy/dx = 1/u · du/dx = (1-x)/(1+x) · du/dx
Find du/dx using the quotient rule: du/dx = [(1-x)(1) - (1+x)(-1)]/(1-x)² du/dx = [1-x + 1+x]/(1-x)² du/dx = 2/(1-x)²
Therefore: dy/dx = (1-x)/(1+x) · 2/(1-x)² dy/dx = 2/[(1+x)(1-x)] dy/dx = 2/(1-x²)
7. A function is given implicitly by x³ + y³ = 3xy. Find dy/dx
Differentiate both sides with respect to x: d/dx(x³ + y³) = d/dx(3xy)
Left side: 3x² + 3y² · dy/dx
Right side (using product rule): 3[x · dy/dx + y · 1] = 3x · dy/dx + 3y
Setting them equal: 3x² + 3y² · dy/dx = 3x · dy/dx + 3y
Collect terms with dy/dx: 3y² · dy/dx - 3x · dy/dx = 3y - 3x dy/dx(3y² - 3x) = 3(y - x) dy/dx = (y - x)/(y² - x)
8. If y = (x² + 1)^(tan⁻¹x), find dy/dx using logarithmic differentiation
Take the natural logarithm of both sides: ln y = ln[(x² + 1)^(tan⁻¹x)] ln y = tan⁻¹x · ln(x² + 1)
Differentiate both sides with respect to x: (1/y) · dy/dx = d/dx[tan⁻¹x · ln(x² + 1)]
Using the product rule on the right side: d/dx[tan⁻¹x · ln(x² + 1)] = tan⁻¹x · d/dx[ln(x² + 1)] + ln(x² + 1) · d/dx[tan⁻¹x]
= tan⁻¹x · (2x)/(x² + 1) + ln(x² + 1) · 1/(1 + x²)
= (2x tan⁻¹x)/(x² + 1) + ln(x² + 1)/(1 + x²)
Therefore: (1/y) · dy/dx = (2x tan⁻¹x + ln(x² + 1))/(x² + 1)
Multiply both sides by y = (x² + 1)^(tan⁻¹x):
dy/dx = (x² + 1)^(tan⁻¹x) · (2x tan⁻¹x + ln(x² + 1))/(x² + 1)
0 Comment to "solution 6 ,7 and 8"
Post a Comment
Note: Only a member of this blog may post a comment.