What is the derivative of f(x)=2+31n(x-4x+1) ?

The derivative of f(x)=2+31n(x-4x+1) is -93n

What is the first derivative of f(x)=2+31n(x-4x+1) ?

The first derivative of f(x)=2+31n(x-4x+1) is -93n

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derivative of

f (x) = 2 + 31 n (x - 4 x + 1)

- 93 n

Solution steps

\frac{d}{d x} (2 + 31 n (x - 4 x + 1))

Simplify

2 + 31 n (x - 4 x + 1) : 2 + 31 n (- 3 x + 1)

= \frac{d}{d x} (2 + 31 n (- 3 x + 1))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2) + \frac{d}{d x} (31 n (- 3 x + 1))

\frac{d}{d x} (2) = 0

\frac{d}{d x} (31 n (- 3 x + 1)) = - 93 n

= 0 - 93 n

Simplify

= - 93 n

(e^{x})^{\frac{1}{2}}

cot^{4} (x)

y = 6 x^{- 2}

\frac{1}{( z - 1 )}

y = 25

What is the derivative of f(x)=2+31n(x-4x+1) ?
The derivative of f(x)=2+31n(x-4x+1) is -93n
What is the first derivative of f(x)=2+31n(x-4x+1) ?
The first derivative of f(x)=2+31n(x-4x+1) is -93n