Solve 1/1+t^2*1/3t^2-6t

$$\frac { 1 } { 1 + t ^ { 2 } } \cdot \frac { 1 } { 3 t ^ { 2 } - 6 t }$$

Evaluate

$$\frac{1}{3t\left(t-2\right)\left(t^{2}+1\right)}$$

Solution Steps

Multiply \frac{1}{1+t^{2}} times \frac{1}{3t^{2}-6t} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{\left(1+t^{2}\right)\left(3t^{2}-6t\right)}$$

Use the distributive property to multiply 1+t^{2} by 3t^{2}-6t.

$$\frac{1}{3t^{2}-6t+3t^{4}-6t^{3}}$$

Differentiate w.r.t. t

$$-\frac{2\left(2t^{3}-3t^{2}+t-1\right)}{3\left(t\left(t-2\right)\left(t^{2}+1\right)\right)^{2}}$$