Find the derivative of y with respect y=(5-5theta)tanh^(-1)theta

Apply Product Rule: Apply the product rule for differentiation. The 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. Let u = (5 - 5theta) and v = tanh^(-1)theta. Then, dy/dtheta = du/dtheta * v + u * dv/dtheta.Differentiate u: Differentiate u = (5 - 5theta) with respect to theta. The derivative of a constant is 0, and the derivative of -5theta with respect to theta is -5. So, du/dtheta = -5.Differentiate v: Differentiate v = tanh^(-1)theta with respect to theta. The derivative of tanh^(-1)theta with respect to theta is 1/(1 - theta^2), due to the inverse hyperbolic tangent derivative formula. So, dv/dtheta = 1/(1 - theta^2).Substitute Derivatives: Substitute the derivatives back into the product rule formula. dy/dtheta = (-5) * tanh^(-1)theta + (5 - 5theta) * (1/(1 - theta^2)).Simplify Expression: Simplify the expression. dy/dtheta = -5 * tanh^(-1)theta + (5/(1 - theta^2)) - (5theta/(1 - theta^2)).

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Find the derivative of $y$ with respect to $\theta$ : $y = (5 - 5\theta)\tanh^{-1}(\theta)$

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Algebra 2

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Q. Find the derivative of

y

with respect to

\theta

y = (5 - 5\theta)\tanh^{-1}(\theta)

Apply Product Rule: Apply the product rule for differentiation. $\newline$ The 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. $\newline$ Let $u = (5 - 5\theta)$ and $v = \tanh^{-1}\theta$ . $\newline$ Then, $\frac{dy}{d\theta} = \frac{du}{d\theta} \cdot v + u \cdot \frac{dv}{d\theta}$ .
Differentiate $u$ : Differentiate $u = (5 - 5\theta)$ with respect to $\theta$ . The derivative of a constant is $0$ , and the derivative of $-5\theta$ with respect to $\theta$ is $-5$ . So, $\frac{du}{d\theta} = -5$ .
Differentiate $v$ : Differentiate $v = \tanh^{-1}\theta$ with respect to $\theta$ . The derivative of $\tanh^{-1}\theta$ with respect to $\theta$ is $\frac{1}{1 - \theta^2}$ , due to the inverse hyperbolic tangent derivative formula. So, $\frac{dv}{d\theta} = \frac{1}{1 - \theta^2}$ .
Substitute Derivatives: Substitute the derivatives back into the product rule formula. $\newline$ $\frac{dy}{d\theta} = (-5) \cdot \tanh^{-1}(\theta) + (5 - 5\theta) \cdot \left(\frac{1}{1 - \theta^2}\right).$
Simplify Expression: Simplify the expression. $\frac{dy}{d\theta} = -5 \cdot \tanh^{-1}(\theta) + \left(\frac{5}{1 - \theta^2}\right) - \left(\frac{5\theta}{1 - \theta^2}\right)$ .