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 + (1+x)²]/(1-x)² = [1-2x+x² + 1+2x+x²]/(1-x)² = (2+2x²)/(1-x)² = 2(1+x²)/(1-x)²

Therefore: dy/dx = 1/(1+x²)

4. If x² + y² = 25, find dy/dx using implicit differentiation

Differentiate both sides with respect to x: d/dx(x² + y²) = d/dx(25) 2x + 2y(dy/dx) = 0

Solve for dy/dx: 2y(dy/dx) = -2x dy/dx = -x/y

5. If y = e^(sin x) · cos x, find dy/dx

Using the product rule: d/dx(uv) = u'v + uv'

Let u = e^(sin x) and v = cos x

Find u': u' = e^(sin x) · d/dx(sin x) = e^(sin x) · cos x

Find v': v' = -sin x

Apply the product rule: dy/dx = u'v + uv' dy/dx = e^(sin x) · cos x · cos x + e^(sin x) · (-sin x) dy/dx = e^(sin x) · cos²x - e^(sin x) · sin x dy/dx = e^(sin x)(cos²x - sin x)

Share this

p L