What is the derivative of at^m+bt^{m+n} ?

The derivative of at^m+bt^{m+n} is 0

What is the first derivative of at^m+bt^{m+n} ?

The first derivative of at^m+bt^{m+n} is 0

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\frac{d}{d x} (a t^{m} + b t^{m + n})

0

Solution steps

\frac{d}{d x} (a t^{m} + b t^{m + n})

Treat

a, m, b, n, t

as a constant

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

\frac{d}{d x} (sin (x) sin (3 x))

4 x^{2} - 2 x + 4

\frac{\partial}{\partial x} (3 x^{2} + 6 x y^{3})

7 - 4 t

\int_{0}^{1} 46 \frac{1}{x ^{p}} d x

What is the derivative of at^m+bt^{m+n} ?
The derivative of at^m+bt^{m+n} is 0
What is the first derivative of at^m+bt^{m+n} ?
The first derivative of at^m+bt^{m+n} is 0