What is the d/(dt)(ln(y^'(t))) ?

The d/(dt)(ln(y^'(t))) is ('t\frac{dy)/(dt)+y}{ty}

What is the first d/(dt)(ln(y^'(t))) ?

The first d/(dt)(ln(y^'(t))) is ('t\frac{dy)/(dt)+y}{ty}

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Popular >

\frac{d}{d t} (ln (y^{^{'}} (t)))

\frac{' t \frac{d y}{d t} + y}{t y}

Solution steps

\frac{d}{d t} (ln (y^{^{'}} (t)))

Apply the chain rule:

\frac{1}{y ^{^{'}} t} \frac{d}{d t} (y^{^{'}} t)

= \frac{1}{y ^{^{'}} t} \frac{d}{d t} (y^{^{'}} t)

\frac{d}{d t} (y^{^{'}} t) =^{'} t \frac{d y}{d t} y^{^{'} - 1} + y^{^{'}}

= \frac{1}{y ^{^{^{'}}} t} (^{^{'}} t \frac{d y}{d t} y^{^{^{'}} - 1} + y^{^{^{'}}})

Simplify

\frac{1}{y ^{^{'}} t} (^{'} t \frac{d y}{d t} y^{^{'} - 1} + y^{^{'}}) : \frac{^{'} t \frac{d y}{d t} + y}{t y}

= \frac{^{'} t \frac{d y}{d t} + y}{t y}

\infty - \infty - \infty

(1 + h)^{3} - 3 - (1^{3} - 4)

r e^{t} + 2 (e^{r t})

A e^{4 x} + B x e^{4 x}

2^{2} - x - 2

What is the d/(dt)(ln(y^'(t))) ?
The d/(dt)(ln(y^'(t))) is ('t\frac{dy)/(dt)+y}{ty}
What is the first d/(dt)(ln(y^'(t))) ?
The first d/(dt)(ln(y^'(t))) is ('t\frac{dy)/(dt)+y}{ty}