What is the derivative of f(t)=(1t^2+0t+0)^{-8} ?

The derivative of f(t)=(1t^2+0t+0)^{-8} is -(16)/(t^{17)}

What is the first derivative of f(t)=(1t^2+0t+0)^{-8} ?

The first derivative of f(t)=(1t^2+0t+0)^{-8} is -(16)/(t^{17)}

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derivative of

f (t) = (1 t^{2} + 0 t + 0)^{- 8}

- \frac{16}{t ^{17}}

Solution steps

\frac{d}{d t} ((1 \cdot t^{2} + 0 \cdot t + 0)^{- 8})

Apply the chain rule:

- \frac{8}{( 1 \cdot t ^{2} + 0 \cdot t + 0 ) ^{9}} \frac{d}{d t} (1 \cdot t^{2} + 0 \cdot t + 0)

= - \frac{8}{( 1 \cdot t ^{2} + 0 \cdot t + 0 ) ^{9}} \frac{d}{d t} (1 \cdot t^{2} + 0 \cdot t + 0)

\frac{d}{d t} (1 \cdot t^{2} + 0 \cdot t + 0) = 2 t

= - \frac{8}{( 1 \cdot t ^{2} + 0 \cdot t + 0 ) ^{9}} \cdot 2 t

Simplify

- \frac{8}{( 1 \cdot t ^{2} + 0 \cdot t + 0 ) ^{9}} \cdot 2 t : - \frac{16}{t ^{17}}

= - \frac{16}{t ^{17}}

\frac{d}{d x} (4^{9^{x^{2}}})

n = 1 \sum \infty (- 1)^{n} x^{2 n}

\frac{1}{x} \cdot \frac{10}{x ^{2} + x + 10}

\frac{d}{d x} (sin (x^{2} + y^{2}))

y^{2} = 4 x, x = 1, x = 4

What is the derivative of f(t)=(1t^2+0t+0)^{-8} ?
The derivative of f(t)=(1t^2+0t+0)^{-8} is -(16)/(t^{17)}
What is the first derivative of f(t)=(1t^2+0t+0)^{-8} ?
The first derivative of f(t)=(1t^2+0t+0)^{-8} is -(16)/(t^{17)}