What is the derivative of y=(1+sin(7t))^{-6} ?

The derivative of y=(1+sin(7t))^{-6} is -(42cos(7t))/((1+sin(7t))^7)

What is the first derivative of y=(1+sin(7t))^{-6} ?

The first derivative of y=(1+sin(7t))^{-6} is -(42cos(7t))/((1+sin(7t))^7)

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derivative of

y = (1 + sin (7 t))^{- 6}

- \frac{42 cos ( 7 t )}{( 1 + sin ( 7 t ) ) ^{7}}

Solution steps

\frac{d}{d t} ((1 + sin (7 t))^{- 6})

Apply the chain rule:

- \frac{6}{( 1 + sin ( 7 t ) ) ^{7}} \frac{d}{d t} (1 + sin (7 t))

= - \frac{6}{( 1 + sin ( 7 t ) ) ^{7}} \frac{d}{d t} (1 + sin (7 t))

\frac{d}{d t} (1 + sin (7 t)) = 7 cos (7 t)

= - \frac{6}{( 1 + sin ( 7 t ) ) ^{7}} \cdot 7 cos (7 t)

Simplify

- \frac{6}{( 1 + sin ( 7 t ) ) ^{7}} \cdot 7 cos (7 t) : - \frac{42 cos ( 7 t )}{( 1 + sin ( 7 t ) ) ^{7}}

= - \frac{42 cos ( 7 t )}{( 1 + sin ( 7 t ) ) ^{7}}

\frac{d y}{d t} = \frac{t e ^{t}}{y 9 + y ^{2}}

\frac{d}{d x} (y^{2} e^{x y - 3})

f (x) = 4 x y

(1 + x^{2}) d y = (y - cot^{- 1} (x)) d x

\int \frac{2}{1 + x} d x

What is the derivative of y=(1+sin(7t))^{-6} ?
The derivative of y=(1+sin(7t))^{-6} is -(42cos(7t))/((1+sin(7t))^7)
What is the first derivative of y=(1+sin(7t))^{-6} ?
The first derivative of y=(1+sin(7t))^{-6} is -(42cos(7t))/((1+sin(7t))^7)