What is the derivative of y=t(ln(4t))^2 ?

The derivative of y=t(ln(4t))^2 is ln^2(4t)+2ln(4t)

What is the first derivative of y=t(ln(4t))^2 ?

The first derivative of y=t(ln(4t))^2 is ln^2(4t)+2ln(4t)

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derivative of

y = t (ln (4 t))^{2}

ln^{2} (4 t) + 2 ln (4 t)

Solution steps

\frac{d}{d t} (t (ln (4 t))^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = (ln (4 t))^{2}

= \frac{d t}{d t} (ln (4 t))^{2} + \frac{d}{d t} ((ln (4 t))^{2}) t

\frac{d t}{d t} = 1

\frac{d}{d t} ((ln (4 t))^{2}) = \frac{2 ln ( 4 t )}{t}

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

Simplify

1 \cdot ln^{2} (4 t) + \frac{2 ln ( 4 t )}{t} t : ln^{2} (4 t) + 2 ln (4 t)

= ln^{2} (4 t) + 2 ln (4 t)

h \to 0 lim (\frac{2 ( x + h ) - 2 x}{h})

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What is the derivative of y=t(ln(4t))^2 ?
The derivative of y=t(ln(4t))^2 is ln^2(4t)+2ln(4t)
What is the first derivative of y=t(ln(4t))^2 ?
The first derivative of y=t(ln(4t))^2 is ln^2(4t)+2ln(4t)