Solve f(x)=2- 5tg root rad x

Evaluate

f (x) = 2 - 5 t g r a d x

Take the derivative of both sides

f^{'} (x) = \frac{d}{d x} (2 - 5 t g r a d x)

Calculate

f^{'} (x) = \frac{d}{d x} (2 - 5 t g r^{\frac{1}{2}} a^{\frac{1}{2}} d^{\frac{1}{2}} x^{\frac{1}{2}})

Use differentiation rule \frac{d}{d x} (f (x) \pm g (x)) = \frac{d}{d x} (f (x)) \pm \frac{d}{d x} (g (x))

f^{'} (x) = \frac{d}{d x} (2) + \frac{d}{d x} (- 5 t g r^{\frac{1}{2}} a^{\frac{1}{2}} d^{\frac{1}{2}} x^{\frac{1}{2}})

Use \frac{d}{d x} (c) = 0 to find derivative

f^{'} (x) = 0 + \frac{d}{d x} (- 5 t g r^{\frac{1}{2}} a^{\frac{1}{2}} d^{\frac{1}{2}} x^{\frac{1}{2}})

Calculate

More Steps

f^{'} (x) = 0 - \frac{5 t g r ^{\frac{1}{2}} a ^{\frac{1}{2}} d ^{\frac{1}{2}}}{2 x ^{\frac{1}{2}}}

Removing 0 doesn't change the value,so remove it from the expression

f^{'} (x) = - \frac{5 t g r ^{\frac{1}{2}} a ^{\frac{1}{2}} d ^{\frac{1}{2}}}{2 x ^{\frac{1}{2}}}

Solution

f^{'} (x) = - \frac{5 t g r \times a \times d}{2 x}

Function