Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.15. y = 2√t tanh(√t)

Accepted Answer

Identify the function to differentiate: $$y = 2 \sqrt{t} 	anh(\sqrt{t}). $$Notice that $$y is a $$product of two functions of $$t$$: $$2 \sqrt{t}$$ and $$	anh(\sqrt{t}). $$Recall the product rule for derivatives: if $$y = u(t) v(t)$$, then $$\frac{dy}{dt} = u'(t) v(t) + u(t) v'(t). $$Here, let $$u(t) = 2 \sqrt{t}$$ and $$v(t) = 	anh(\sqrt{t}). $$Find $$u'(t)$$: Since $$u(t) = 2 t$$^{1/2}$$, use the power rule to get u$$'(t) = 2 	imes \frac{1}{2} $$t^{-1/2}$$ = $$t^{-1/2}. $$Find $$v'(t)$$: Since $$v(t) = 	anh(\sqrt{t})$$, apply the chain rule. First, recall that $$\frac{d}{dx} 	anh(x) = \operatorname{sech}^2(x). $$Let $$g(t) = \sqrt{t} = t$$^{1/2}$$, so v(t$$) = 	anh(g(t))$$. Then, v$$'(t) = \operatorname{sech}^2(g(t)) \cdot g'(t)$$, where g$$'(t) = \frac{1}{2} $$t^{-1/2}. $$Combine all parts using the product rule: $$\frac{dy}{dt} = u'(t) v(t) + u(t) v'(t) = t$$^{-1/2}$$ 	anh(\sqrt{t}) + 2 \sqrt{t} 	imes \operatorname{sech}^2(\sqrt{t}) 	imes \frac{1}{2} t$$^{-1/2}$$. Simplify the expression as needed.

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
15. y = 2√t tanh(√t)

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Key Concepts

Derivative of Composite Functions (Chain Rule)

Derivative of Hyperbolic Functions

Product Rule for Differentiation