What is the derivative of y=ln(t(t^2+3)^3) ?

The derivative of y=ln(t(t^2+3)^3) is (7t^2+3)/(t(t^2+3))

What is the first derivative of y=ln(t(t^2+3)^3) ?

The first derivative of y=ln(t(t^2+3)^3) is (7t^2+3)/(t(t^2+3))

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derivative of

y = ln (t (t^{2} + 3)^{3})

\frac{7 t ^{2} + 3}{t ( t ^{2} + 3 )}

Solution steps

\frac{d}{d t} (ln (t (t^{2} + 3)^{3}))

Apply the chain rule:

\frac{1}{t ( t ^{2} + 3 ) ^{3}} \frac{d}{d t} (t (t^{2} + 3)^{3})

= \frac{1}{t ( t ^{2} + 3 ) ^{3}} \frac{d}{d t} (t (t^{2} + 3)^{3})

\frac{d}{d t} (t (t^{2} + 3)^{3}) = (t^{2} + 3)^{3} + 6 t^{2} (t^{2} + 3)^{2}

= \frac{1}{t ( t ^{2} + 3 ) ^{3}} ((t^{2} + 3)^{3} + 6 t^{2} (t^{2} + 3)^{2})

Simplify

\frac{1}{t ( t ^{2} + 3 ) ^{3}} ((t^{2} + 3)^{3} + 6 t^{2} (t^{2} + 3)^{2}) : \frac{7 t ^{2} + 3}{t ( t ^{2} + 3 )}

= \frac{7 t ^{2} + 3}{t ( t ^{2} + 3 )}

What is the derivative of y=ln(t(t^2+3)^3) ?
The derivative of y=ln(t(t^2+3)^3) is (7t^2+3)/(t(t^2+3))
What is the first derivative of y=ln(t(t^2+3)^3) ?
The first derivative of y=ln(t(t^2+3)^3) is (7t^2+3)/(t(t^2+3))