Solve mn+2mn(m)-3(-2m^2n)+61

$$mn + 2 mn ( m ) - 3 ( - 2 m ^ { 2 } n ) + 61$$

Differentiate w.r.t. m

$$16mn+n$$

Steps Using Derivative Rule for Sum

Multiply m and m to get m^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}m}(mn+2m^{2}n-3\left(-2\right)m^{2}n+61)$$

Multiply 3 and -2 to get -6.

$$\frac{\mathrm{d}}{\mathrm{d}m}(mn+2m^{2}n-\left(-6m^{2}n\right)+61)$$

The opposite of -6m^{2}n is 6m^{2}n.

$$\frac{\mathrm{d}}{\mathrm{d}m}(mn+2m^{2}n+6m^{2}n+61)$$

Combine 2m^{2}n and 6m^{2}n to get 8m^{2}n.

$$\frac{\mathrm{d}}{\mathrm{d}m}(mn+8m^{2}n+61)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2\times 8nm^{2-1}+nm^{1-1}$$

Multiply 2 times 8n.

$$16nm^{2-1}+nm^{1-1}$$

Subtract 1 from 2.

$$16nm^{1}+nm^{1-1}$$

Subtract 1 from 1.

$$16nm^{1}+nm^{0}$$

For any term t, t^{1}=t.

$$16nm+nm^{0}$$

For any term t except 0, t^{0}=1.

$$16nm+n\times 1$$

For any term t, t\times 1=t and 1t=t.

$$16nm+n$$

Evaluate

$$mn+8nm^{2}+61$$