What is the d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) ?

The d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) is (e^{(5t)/4}(5(t^2+1)-2t))/(4(t^2+1)^{5/4)}

What is the first d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) ?

The first d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) is (e^{(5t)/4}(5(t^2+1)-2t))/(4(t^2+1)^{5/4)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1})

= \frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1})

\frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1}) = \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}}

= \frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \cdot \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \cdot \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}} : \frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

= \frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

\frac{y ^{^{''}}}{y} = - 3

\int (x^{2} \cdot sin (3 x)) d x

\frac{d y}{d x} + 10 x^{4} y = 50 x^{9} e^{3 x^{5}}

\int 1 + t^{2} d t

\int - 25 sin (- 5 t) d t

What is the d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) ?
The d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) is (e^{(5t)/4}(5(t^2+1)-2t))/(4(t^2+1)^{5/4)}
What is the first d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) ?
The first d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)}) is (e^{(5t)/4}(5(t^2+1)-2t))/(4(t^2+1)^{5/4)}