What is the d/(dm)(16m^2+40m+25) ?

The d/(dm)(16m^2+40m+25) is 32m+40

What is the first d/(dm)(16m^2+40m+25) ?

The first d/(dm)(16m^2+40m+25) is 32m+40

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\frac{d}{d m} (16 m^{2} + 40 m + 25)

32 m + 40

Solution steps

\frac{d}{d m} (16 m^{2} + 40 m + 25)

Apply the Sum/Difference Rule:

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

= \frac{d}{d m} (16 m^{2}) + \frac{d}{d m} (40 m) + \frac{d}{d m} (25)

\frac{d}{d m} (16 m^{2}) = 32 m

\frac{d}{d m} (40 m) = 40

\frac{d}{d m} (25) = 0

= 32 m + 40 + 0

Simplify

= 32 m + 40

\frac{d}{d x} (3 x^{2} sin (5 x))

\int \frac{2 x - 3}{( x ^{2} - 3 x + 6 ) ^{2}} d x

\int tan (x) - x d x

x \to 4 lim (\frac{x}{x ^{2} + 4})

arctan (3 x)

What is the d/(dm)(16m^2+40m+25) ?
The d/(dm)(16m^2+40m+25) is 32m+40
What is the first d/(dm)(16m^2+40m+25) ?
The first d/(dm)(16m^2+40m+25) is 32m+40