Solve -01x^2+0.9x

$$- 01 x ^ { 2 } + 0.9 x$$

Evaluate

$$\frac{9x}{10}$$

Solution Steps

Multiply 0 and 1 to get 0.

$$0x^{2}+0.9x$$

Anything times zero gives zero.

$$0+0.9x$$

Anything plus zero gives itself.

$$0.9x$$

Differentiate w.r.t. x

$$0.9$$

Steps Using Definition of a Derivative

Multiply 0 and 1 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(0x^{2}+0.9x)$$

Anything times zero gives zero.

$$\frac{\mathrm{d}}{\mathrm{d}x}(0+0.9x)$$

Anything plus zero gives itself.

$$\frac{\mathrm{d}}{\mathrm{d}x}(0.9x)$$

The derivative of ax^{n} is nax^{n-1}.

$$0.9x^{1-1}$$

Subtract 1 from 1.

$$0.9x^{0}$$

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

$$0.9\times 1$$

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

$$0.9$$