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Find the derivative of yy withy=(5−5θ)tanh⁡−1(θ)y=(5-5\theta)\tanh^{-1}(\theta)

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Full solution

Q. Find the derivative of yy withy=(5−5θ)tanh⁡−1(θ)y=(5-5\theta)\tanh^{-1}(\theta)
  1. Apply Product Rule: Apply the product rule for differentiation.\newlineThe product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.\newlineLet u=(5−5θ)u = (5 - 5\theta) and v=tanh⁡−1(θ)v = \tanh^{-1}(\theta), then y=u⋅vy = u \cdot v.\newlineWe need to find u′u' (the derivative of uu with respect to θ\theta) and v′v' (the derivative of vv with respect to θ\theta).
  2. Differentiate uu: Differentiate u=(5−5θ)u = (5 - 5\theta) with respect to θ\theta. The derivative of a constant is 00, and the derivative of −5θ-5\theta with respect to θ\theta is −5-5. So, u′=ddθ(5−5θ)=0−5=−5u' = \frac{d}{d\theta} (5 - 5\theta) = 0 - 5 = -5.
  3. Differentiate vv: Differentiate v=tanh⁡−1(θ)v = \tanh^{-1}(\theta) with respect to θ\theta. The derivative of the inverse hyperbolic tangent function tanh⁡−1(x)\tanh^{-1}(x) with respect to xx is 11−x2\frac{1}{1 - x^2}. So, v′=ddθ(tanh⁡−1(θ))=11−θ2v' = \frac{d}{d\theta} (\tanh^{-1}(\theta)) = \frac{1}{1 - \theta^2}.
  4. Apply Product Rule: Apply the product rule using the derivatives from steps 22 and 33.\newlineUsing the product rule, y′=u′⋅v+u⋅v′y' = u' \cdot v + u \cdot v'.\newlineSubstitute u′=−5u' = -5 and v′=11−θ2v' = \frac{1}{1 - \theta^2} into the equation.\newliney′=(−5)⋅tanh⁡−1(θ)+(5−5θ)⋅(11−θ2)y' = (-5) \cdot \tanh^{-1}(\theta) + (5 - 5\theta) \cdot \left(\frac{1}{1 - \theta^2}\right).
  5. Simplify Expression: Simplify the expression for y′y'. Combine like terms and simplify the expression if possible. y′=−5⋅tanh⁡−1(θ)+(51−θ2)−(5θ1−θ2).y' = -5 \cdot \tanh^{-1}(\theta) + \left(\frac{5}{1 - \theta^2}\right) - \left(\frac{5\theta}{1 - \theta^2}\right).

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