What is the derivative of-3^{-2}(tan(7x)^5) ?

The derivative of-3^{-2}(tan(7x)^5) is -35/9 tan^4(7x)sec^2(7x)

What is the first derivative of-3^{-2}(tan(7x)^5) ?

The first derivative of-3^{-2}(tan(7x)^5) is -35/9 tan^4(7x)sec^2(7x)

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\frac{d}{d x} (- 3^{- 2} (tan (7 x))^{5})

- \frac{35}{9} tan^{4} (7 x) sec^{2} (7 x)

Solution steps

\frac{d}{d x} (- 3^{- 2} (tan (7 x))^{5})

Take the constant out:

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

= - 3^{- 2} \frac{d}{d x} ((tan (7 x))^{5})

Apply the chain rule:

5 (tan (7 x))^{4} \frac{d}{d x} (tan (7 x))

= 5 (tan (7 x))^{4} \frac{d}{d x} (tan (7 x))

\frac{d}{d x} (tan (7 x)) = sec^{2} (7 x) \cdot 7

= - 3^{- 2} \cdot 5 (tan (7 x))^{4} sec^{2} (7 x) \cdot 7

Simplify

- 3^{- 2} \cdot 5 tan^{4} (7 x) sec^{2} (7 x) \cdot 7 : - \frac{35}{9} tan^{4} (7 x) sec^{2} (7 x)

= - \frac{35}{9} tan^{4} (7 x) sec^{2} (7 x)

(2 cos (x))^{^{'}}

y^{2} = x - 2, y = x - 4

\int (sec (2 x) + tan (2 x)) d x

f (x) = \frac{2 t + 1}{t + 3}

y = 2 x^{2} + 3, (0, 3)

What is the derivative of-3^{-2}(tan(7x)^5) ?
The derivative of-3^{-2}(tan(7x)^5) is -35/9 tan^4(7x)sec^2(7x)
What is the first derivative of-3^{-2}(tan(7x)^5) ?
The first derivative of-3^{-2}(tan(7x)^5) is -35/9 tan^4(7x)sec^2(7x)