What is the derivative of ccos(t+t^2,t) ?

The derivative of ccos(t+t^2,t) is 0

What is the first derivative of ccos(t+t^2,t) ?

The first derivative of ccos(t+t^2,t) is 0

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\frac{d}{d x} (c cos (t) + t^{2}, t)

0

Solution steps

\frac{d}{d x} (c cos (t) + t^{2}, t)

Treat

c, t

as a constant

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

\frac{d}{d x} (arcsec (e^{3 x}))

y^{^{'}} = \frac{sec ( y )}{( x + 2 ) ^{2}}

f (x) = - 3 + (x + 4)^{2}, at x = - 5

\frac{2 \cdot ( s ^{2} + 1 )}{s ^{4}}

\frac{d}{d x} (e^{\frac{x ^{2}}{2 a}})

What is the derivative of ccos(t+t^2,t) ?
The derivative of ccos(t+t^2,t) is 0
What is the first derivative of ccos(t+t^2,t) ?
The first derivative of ccos(t+t^2,t) is 0