What is the derivative of y=565+t(37t^{0.6}-103) ?

The derivative of y=565+t(37t^{0.6}-103) is 59.2t^{0.6}-103

What is the first derivative of y=565+t(37t^{0.6}-103) ?

The first derivative of y=565+t(37t^{0.6}-103) is 59.2t^{0.6}-103

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derivative of

y = 565 + t (37 t^{0.6} - 103)

59.2 t^{0.6} - 103

Solution steps

\frac{d}{d t} (565 + t (37 t^{0.6} - 103))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (565) + \frac{d}{d t} (t (37 t^{0.6} - 103))

\frac{d}{d t} (565) = 0

\frac{d}{d t} (t (37 t^{0.6} - 103)) = 59.2 t^{0.6} - 103

= 0 + 59.2 t^{0.6} - 103

Simplify

= 59.2 t^{0.6} - 103

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x e^{- 2 x}

\frac{d q}{d t} = k q^{2}

f (x) = (3 x + 1)^{5}

\int \frac{2 x ^{4} + 8 x ^{3} - x}{x ^{3}} d x

What is the derivative of y=565+t(37t^{0.6}-103) ?
The derivative of y=565+t(37t^{0.6}-103) is 59.2t^{0.6}-103
What is the first derivative of y=565+t(37t^{0.6}-103) ?
The first derivative of y=565+t(37t^{0.6}-103) is 59.2t^{0.6}-103