What is the derivative of bcos(t)+t^2sin(t) ?

The derivative of bcos(t)+t^2sin(t) is cos(t)

What is the first derivative of bcos(t)+t^2sin(t) ?

The first derivative of bcos(t)+t^2sin(t) is cos(t)

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derivative of

b cos (t) + t^{2} sin (t)

cos (t)

Solution steps

\frac{d}{d b} (b cos (t) + t^{2} sin (t))

Treat

t

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d b} (b cos (t)) + \frac{d}{d b} (t^{2} sin (t))

\frac{d}{d b} (b cos (t)) = cos (t)

\frac{d}{d b} (t^{2} sin (t)) = 0

= cos (t) + 0

Simplify

= cos (t)

\int_{- 1}^{1} 3 x^{2} x^{3} + 2 d x

\frac{\partial}{\partial x} (4 x y - x^{4} - y^{4})

y = x^{3}, y = 2 x - x^{2}

x \to 2 lim (\frac{( x ^{2} + 7 x - 44 )}{x ^{2} - 6 x + 8})

What is the derivative of bcos(t)+t^2sin(t) ?
The derivative of bcos(t)+t^2sin(t) is cos(t)
What is the first derivative of bcos(t)+t^2sin(t) ?
The first derivative of bcos(t)+t^2sin(t) is cos(t)