What is the d/(dt)(tan(t+pi/4)) ?

The d/(dt)(tan(t+pi/4)) is sec^2(t+pi/4)

What is the first d/(dt)(tan(t+pi/4)) ?

The first d/(dt)(tan(t+pi/4)) is sec^2(t+pi/4)

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\frac{d}{d t} (tan (t + \frac{π}{4}))

sec^{2} (t + \frac{π}{4})

Solution steps

\frac{d}{d t} (tan (t + \frac{π}{4}))

Apply the chain rule:

sec^{2} (t + \frac{π}{4}) \frac{d}{d t} (t + \frac{π}{4})

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

\frac{d}{d t} (t + \frac{π}{4}) = 1

= sec^{2} (t + \frac{π}{4}) \cdot 1

Simplify

= sec^{2} (t + \frac{π}{4})

\frac{d y}{d x} = y - y^{5}

\frac{d ^{2} y}{d x ^{2}} - 2 \cdot \frac{d y}{d x} + a y = 0, a > 1

\frac{x ^{5} + 1}{x}

\int_{0}^{1} \frac{8 y}{e ^{2 y}} d y

\int_{0}^{1} 2 π x (x^{3} - x^{4}) d x

What is the d/(dt)(tan(t+pi/4)) ?
The d/(dt)(tan(t+pi/4)) is sec^2(t+pi/4)
What is the first d/(dt)(tan(t+pi/4)) ?
The first d/(dt)(tan(t+pi/4)) is sec^2(t+pi/4)