What is the derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 ?

The derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 is -0.11(t+1.1)^{1.2}+0.7

What is the first derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 ?

The first derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 is -0.11(t+1.1)^{1.2}+0.7

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derivative of

- 0.05 (t + 1.1)^{2.2} + 0.7 t + 0.9

- 0.11 (t + 1.1)^{1.2} + 0.7

Solution steps

\frac{d}{d t} (- 0.05 (t + 1.1)^{2.2} + 0.7 t + 0.9)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d t} (0.05 (t + 1.1)^{2.2}) + \frac{d}{d t} (0.7 t) + \frac{d}{d t} (0.9)

\frac{d}{d t} (0.05 (t + 1.1)^{2.2}) = 0.11 (t + 1.1)^{1.2}

\frac{d}{d t} (0.7 t) = 0.7

\frac{d}{d t} (0.9) = 0

= - 0.11 (t + 1.1)^{1.2} + 0.7 + 0

Simplify

= - 0.11 (t + 1.1)^{1.2} + 0.7

\int \frac{1}{( y - 2 ) ( y + 2 )} d y

\frac{d}{d x} (x - x)

\int e^{x} (3 e^{x} + 1)^{- \frac{3}{2}} d x

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y^{^{''}} + (\frac{2}{x}) = 0

What is the derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 ?
The derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 is -0.11(t+1.1)^{1.2}+0.7
What is the first derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 ?
The first derivative of-0.05(t+1.1)^{2.2}+0.7t+0.9 is -0.11(t+1.1)^{1.2}+0.7