How to find instantaneous rate of change for 4 t + 6 when t=6?

the instantaneous value is the derivative of f(t) at x = adifferentiate using the chain rulegiven y=f(g(x)) then d y d x = f ′ ( g ( x ) ) × g ′ ( x ) ← chain rulegiven y=f(g(x)) then d y d x = f ′ ( g ( x ) ) × g ′ ( x ) ← chain rule f ( t ) = 4 t + 6 = ( 4 t + 6 ) 1 2 ⇒ f ′ ( t ) = 1 2 ( 4 t + 6 ) − 1 2 × d d t ( 4 t + 6 ) = 2 ( 4 t + 6 ) 1 2 ⇒ f ′ ( 6 ) = 2 30

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High School
Calculus and Analysis
Calculus 1
Derivatives

bandikizaui

Answered question

2022-07-10

How to find instantaneous rate of change for

\sqrt{4 t + 6}

when t=6?

Answer & Explanation

Ordettyreomqu

Beginner2022-07-11Added 22 answers

the instantaneous value is the derivative of f(t) at x = a
differentiate using the chain rule
given y=f(g(x)) then

\frac{d y}{d x} = f^{'} (g (x)) \times g^{'} (x) \leftarrow

chain rule
given y=f(g(x)) then

\frac{d y}{d x} = f^{'} (g (x)) \times g^{'} (x) \leftarrow

chain rule

f (t) = \sqrt{4 t + 6} = (4 t + 6)^{\frac{1}{2}}

\Rightarrow f^{'} (t) = \frac{1}{2} (4 t + 6)^{- \frac{1}{2}} \times \frac{d}{d t} (4 t + 6)

= \frac{2}{(4 t + 6)^{\frac{1}{2}}}

\Rightarrow f^{'} (6) = \frac{2}{\sqrt{30}}

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