Resolved: "Find the derivatives of the following functions. g(t)=sin⁡h−1t" - Plainmath

Cheyanne Leigh

Answered question

2021-04-18

Find the derivatives of the following functions.

g (t) = {\sin h}^{- 1} \sqrt{t}

Answer & Explanation

falhiblesw

Skilled2021-04-20Added 97 answers

Step 1
The given function is,

g (t) = {\sin h}^{- 1} \sqrt{t}

Using the chain rule of differentiation,

g^{'} (t) = \frac{d}{d t} (g (t))

= [\frac{d}{d (\sqrt{t})} (g (t))] \cdot [\frac{d}{d t} \sqrt{t}]

Step 2
Putting the function, we get

g^{'} (t) = [\frac{d}{d (\sqrt{t})} ({\sin h}^{- 1} \sqrt{t})] \cdot [\frac{d}{d t} \sqrt{t}]

= \frac{1}{\sqrt{{(\sqrt{t})}^{2} + 1}} \cdot (\frac{1}{2 \sqrt{t}})

= \frac{1}{2 \sqrt{t} (\sqrt{t + 1})}

Therefore, the derivative of the given function is

\frac{1}{2 \sqrt{t} (\sqrt{t + 1})}

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