Solve alpha+2+(-8)(-3)+4/sqrt{1^2+(-8)^2}

$$\alpha + \frac { 2 + ( - 8 ) ( - 3 ) + 4 } { \sqrt { 1 ^ { 2 } + ( - 8 ) ^ { 2 } } }$$

Evaluate

$$\alpha +\frac{6\sqrt{65}}{13}$$

Solution Steps

Multiply -8 and -3 to get 24.

$$\alpha +\frac{2+24+4}{\sqrt{1^{2}+\left(-8\right)^{2}}}$$

Add 2 and 24 to get 26.

$$\alpha +\frac{26+4}{\sqrt{1^{2}+\left(-8\right)^{2}}}$$

Add 26 and 4 to get 30.

$$\alpha +\frac{30}{\sqrt{1^{2}+\left(-8\right)^{2}}}$$

Calculate 1 to the power of 2 and get 1.

$$\alpha +\frac{30}{\sqrt{1+\left(-8\right)^{2}}}$$

Calculate -8 to the power of 2 and get 64.

$$\alpha +\frac{30}{\sqrt{1+64}}$$

Add 1 and 64 to get 65.

$$\alpha +\frac{30}{\sqrt{65}}$$

Rationalize the denominator of \frac{30}{\sqrt{65}} by multiplying numerator and denominator by \sqrt{65}.

$$\alpha +\frac{30\sqrt{65}}{\left(\sqrt{65}\right)^{2}}$$

The square of \sqrt{65} is 65.

$$\alpha +\frac{30\sqrt{65}}{65}$$

Divide 30\sqrt{65} by 65 to get \frac{6}{13}\sqrt{65}.

$$\alpha +\frac{6}{13}\sqrt{65}$$

Differentiate w.r.t. α

$$1$$