Solution 09

 Given: y = sin⁻¹(2x√(1-x²)) Using the chain rule with d/dx[sin⁻¹(u)] = 1/√(1-u²) · du/dx Let u = 2x√(1-x²) Step 1: Find du/dx using the product rule. u = 2x · √(1-x²) du/dx = 2x · d/dx[√(1-x²)] + √(1-x²) · d/dx[2x] For d/dx[√(1-x²)] = d/dx[(1-x²)^(1/2)] = (1/2)(1-x²)^(-1/2) · (-2x) = -x/√(1-x²) So: du/dx = 2x · (-x/√(1-x²)) + √(1-x²) · 2 du/dx = -2x²/√(1-x²) + 2√(1-x²) du/dx...

solution 6 ,7 and 8

 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...

Solution 3, 4 and 5

 3. If y = tan⁻¹((1+x)/(1-x)), find dy/dx Using the chain rule with the derivative of tan⁻¹(u) = 1/(1+u²) · du/dx Let u = (1+x)/(1-x) First, 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 Now apply the chain rule: dy/dx = 1/(1 + ((1+x)/(1-x))²) · 2/(1-x)² Simplify the denominator: 1 + ((1+x)/(1-x))² = [(1-x)²...

solution 2

 To differentiate y = x^x, I need to use logarithmic differentiation since both the base and exponent contain x. Given: y = x^x Step 1: Take the natural logarithm of both sides. ln y = ln(x^x) ln y = x ln x Step 2: Differentiate both sides with respect to x. d/dx(ln y) = d/dx(x ln x) Step 3: Apply the chain rule on the left side and product rule on the right side. Left side:...

Solution 1

 To find the derivative of y = ln(sin x), I'll use the chain rule. Given: y = ln(sin x) Using the chain rule: dy/dx = d/dx[ln(sin x)] The derivative of ln(u) is 1/u · du/dx, where u = sin x. So: dy/dx = 1/(sin x) · d/dx(sin x) Since d/dx(sin x) = cos x: dy/dx = 1/(sin x) · cos x = cos x/sin x Therefore: dy/dx = cot x We can also write this as cos x/sin x, but cot x is the...

Solution Q. 10

 Given the equation x=etan⁡−1(y−x2x2)x=etan−1(x2y−x2​), we need to find (dydx)(dxdy​) at x=1x=1.First, we take the natural logarithm of both sides:ln⁡x=tan⁡−1(y−x2x2)lnx=tan−1(x2y−x2​)Let u=y−x2x2u=x2y−x2​. Then the equation becomes:ln⁡x=tan⁡−1(u)lnx=tan−1(u)Taking the tangent of both sides, we get:u=tan⁡(ln⁡x)u=tan(lnx)Substituting back u=y−x2x2u=x2y−x2​,...