What is the derivative of y= 1/(7t+3) ?

The derivative of y= 1/(7t+3) is -7/((7t+3)^2)

What is the first derivative of y= 1/(7t+3) ?

The first derivative of y= 1/(7t+3) is -7/((7t+3)^2)

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derivative of

y = \frac{1}{7 t + 3}

- \frac{7}{( 7 t + 3 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1}{7 t + 3})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d t} ((7 t + 3)^{- 1})

Apply the chain rule:

- \frac{1}{( 7 t + 3 ) ^{2}} \frac{d}{d t} (7 t + 3)

= - \frac{1}{( 7 t + 3 ) ^{2}} \frac{d}{d t} (7 t + 3)

\frac{d}{d t} (7 t + 3) = 7

= - \frac{1}{( 7 t + 3 ) ^{2}} \cdot 7

Simplify

- \frac{1}{( 7 t + 3 ) ^{2}} \cdot 7 : - \frac{7}{( 7 t + 3 ) ^{2}}

= - \frac{7}{( 7 t + 3 ) ^{2}}

\int_{0}^{15} \frac{x}{x + 1} d x

- x^{3} + 3, y = x, x = - 1, x = 1

\int (x cos (x) + sin (x)) 1 + u d x

\int x^{3} - 4 x + 2 d x

\frac{d r}{d θ} = - π sin (π θ), r (2) = 3

What is the derivative of y= 1/(7t+3) ?
The derivative of y= 1/(7t+3) is -7/((7t+3)^2)
What is the first derivative of y= 1/(7t+3) ?
The first derivative of y= 1/(7t+3) is -7/((7t+3)^2)