Solve t^0*t^-48/t^-1*t^-6

$$\frac { t ^ { 0 } \cdot t ^ { - 48 } } { t ^ { - 1 } \cdot t ^ { - 6 } }$$

Evaluate

$$\frac{1}{t^{41}}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 0 and -48 to get -48.

$$\frac{t^{-48}}{t^{-1}t^{-6}}$$

To multiply powers of the same base, add their exponents. Add -1 and -6 to get -7.

$$\frac{t^{-48}}{t^{-7}}$$

Rewrite t^{-7} as t^{-48}t^{41}. Cancel out t^{-48} in both numerator and denominator.

$$\frac{1}{t^{41}}$$

Differentiate w.r.t. t

$$-\frac{41}{t^{42}}$$

Steps Using Chain Rule

To multiply powers of the same base, add their exponents. Add 0 and -48 to get -48.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{t^{-48}}{t^{-1}t^{-6}})$$

To multiply powers of the same base, add their exponents. Add -1 and -6 to get -7.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{t^{-48}}{t^{-7}})$$

Rewrite t^{-7} as t^{-48}t^{41}. Cancel out t^{-48} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}t}(\frac{1}{t^{41}})$$

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$-\left(t^{41}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}t}(t^{41})$$

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}.

$$-\left(t^{41}\right)^{-2}\times 41t^{41-1}$$

Simplify.

$$-41t^{40}\left(t^{41}\right)^{-2}$$