What is the derivative of k(sec(x+tan(x))) ?

The derivative of k(sec(x+tan(x))) is -k/(-1+sin(x))

What is the first derivative of k(sec(x+tan(x))) ?

The first derivative of k(sec(x+tan(x))) is -k/(-1+sin(x))

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\frac{d}{d x} (k (sec (x) + tan (x)))

- \frac{k}{- 1 + sin ( x )}

Solution steps

\frac{d}{d x} (k (sec (x) + tan (x)))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= k \frac{d}{d x} (sec (x) + tan (x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= k (\frac{d}{d x} (sec (x)) + \frac{d}{d x} (tan (x)))

\frac{d}{d x} (sec (x)) = sec (x) tan (x)

\frac{d}{d x} (tan (x)) = sec^{2} (x)

= k (sec (x) tan (x) + sec^{2} (x))

Simplify

k (sec (x) tan (x) + sec^{2} (x)) : - \frac{k}{- 1 + sin ( x )}

= - \frac{k}{- 1 + sin ( x )}

(4 + x) y^{^{'}} = 3 y

\int (1 + \frac{9}{4} x)^{\frac{1}{2}} d x

\frac{d y}{d x} = 1 + y

\int sin (x - 3) d x

3 cos (x) + 2 sin (x)

What is the derivative of k(sec(x+tan(x))) ?
The derivative of k(sec(x+tan(x))) is -k/(-1+sin(x))
What is the first derivative of k(sec(x+tan(x))) ?
The first derivative of k(sec(x+tan(x))) is -k/(-1+sin(x))