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Popular >

derivative of 5sec(xtan(x))

Solution

\frac{d}{d x} (5 sec (x) tan (x))

Solution

5 (- sec (x) + 2 sec^{3} (x))

Solution steps

\frac{d}{d x} (5 sec (x) tan (x))

Take the constant out:

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

= 5 \frac{d}{d x} (sec (x) tan (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sec (x), g = tan (x)

= 5 (\frac{d}{d x} (sec (x)) tan (x) + \frac{d}{d x} (tan (x)) sec (x))

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

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

= 5 (sec (x) tan (x) tan (x) + sec^{2} (x) sec (x))

Simplify

5 (sec (x) tan (x) tan (x) + sec^{2} (x) sec (x)) : 5 (- sec (x) + 2 sec^{3} (x))

= 5 (- sec (x) + 2 sec^{3} (x))

Popular Examples

limit as u approaches 8 of ((u^2-5u-24))/(u-8)

u \to 8 lim (\frac{( u ^{2} - 5 u - 24 )}{u - 8})

(\partial)/(\partial x)(3+200x^2+300y^2)

\frac{\partial}{\partial x} (3 + 200 x^{2} + 300 y^{2})

tangent of f(x)=((2x-1))/(x+3),\at x=2 tangent of

f (x) = \frac{( 2 x - 1 )}{x + 3}, at x = 2

derivative of f(x)=(2x-4)^4(x^2+x+1)^5 derivative of

f (x) = (2 x - 4)^{4} (x^{2} + x + 1)^{5}

(\partial)/(\partial x)(y/(2-x))

\frac{\partial}{\partial x} (\frac{y}{2 - x})

Frequently Asked Questions (FAQ)

What is the derivative of 5sec(xtan(x)) ?
The derivative of 5sec(xtan(x)) is 5(-sec(x)+2sec^3(x))
What is the first derivative of 5sec(xtan(x)) ?
The first derivative of 5sec(xtan(x)) is 5(-sec(x)+2sec^3(x))