What is the derivative of log_{10}(1+tan^1(x)) ?

The derivative of log_{10}(1+tan^1(x)) is (sec^2(x))/(ln(10)(1+tan(x)))

What is the first derivative of log_{10}(1+tan^1(x)) ?

The first derivative of log_{10}(1+tan^1(x)) is (sec^2(x))/(ln(10)(1+tan(x)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

lo g_{10} (1 + tan^{1} (x))

\frac{sec ^{2} ( x )}{ln ( 10 ) ( 1 + tan ( x ) )}

Solution steps

\frac{d}{d x} (lo g_{10} (1 + tan^{1} (x)))

Simplify

lo g_{10} (1 + tan^{1} (x)) : lo g_{10} (1 + tan (x))

= \frac{d}{d x} (lo g_{10} (1 + tan (x)))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( 1 + tan ( x ) )}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (1 + tan (x)))

Apply the chain rule:

\frac{1}{1 + tan ( x )} \frac{d}{d x} (1 + tan (x))

= \frac{1}{1 + tan ( x )} \frac{d}{d x} (1 + tan (x))

\frac{d}{d x} (1 + tan (x)) = sec^{2} (x)

= \frac{1}{ln ( 10 )} \cdot \frac{1}{1 + tan ( x )} sec^{2} (x)

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{1 + tan ( x )} sec^{2} (x) : \frac{sec ^{2} ( x )}{ln ( 10 ) ( 1 + tan ( x ) )}

= \frac{sec ^{2} ( x )}{ln ( 10 ) ( 1 + tan ( x ) )}

\frac{d}{d x} (a (b (x - h))^{2} + k)

\frac{d}{d t} (- 60 + 20 t)

y = 5 x + x + 5

sin (2 + x)

x = t^{2} (t - 1)

What is the derivative of log_{10}(1+tan^1(x)) ?
The derivative of log_{10}(1+tan^1(x)) is (sec^2(x))/(ln(10)(1+tan(x)))
What is the first derivative of log_{10}(1+tan^1(x)) ?
The first derivative of log_{10}(1+tan^1(x)) is (sec^2(x))/(ln(10)(1+tan(x)))