What is the derivative of y=tan(t/(1+t^2)) ?

The derivative of y=tan(t/(1+t^2)) is (sec^2(t/(1+t^2))(1-t^2))/((1+t^2)^2)

What is the first derivative of y=tan(t/(1+t^2)) ?

The first derivative of y=tan(t/(1+t^2)) is (sec^2(t/(1+t^2))(1-t^2))/((1+t^2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = tan (\frac{t}{1 + t ^{2}})

\frac{sec ^{2} ( \frac{t}{1 + t ^{2}} ) ( 1 - t ^{2} )}{( 1 + t ^{2} ) ^{2}}

Solution steps

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

Apply the chain rule:

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

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

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

= sec^{2} (\frac{t}{1 + t ^{2}}) \frac{1 - t ^{2}}{( 1 + t ^{2} ) ^{2}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sec ^{2} ( \frac{t}{1 + t ^{2}} ) ( 1 - t ^{2} )}{( 1 + t ^{2} ) ^{2}}

\int_{0}^{1} 4 π (x^{2} - x^{10}) d x

\frac{3}{x}

\int 2 x (x^{2} + 5)^{5} d x

\frac{280}{s ^{2} + 14 s + 245}

\int m + n x d x

What is the derivative of y=tan(t/(1+t^2)) ?
The derivative of y=tan(t/(1+t^2)) is (sec^2(t/(1+t^2))(1-t^2))/((1+t^2)^2)
What is the first derivative of y=tan(t/(1+t^2)) ?
The first derivative of y=tan(t/(1+t^2)) is (sec^2(t/(1+t^2))(1-t^2))/((1+t^2)^2)