Solve k^80*k^-28:k^93

$$k ^ { 80 } \cdot k ^ { - 28 } : k ^ { 93 }$$

Evaluate

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

Solution Steps

To multiply powers of the same base, add their exponents. Add 80 and -28 to get 52.

$$\frac{k^{52}}{k^{93}}$$

Rewrite k^{93} as k^{52}k^{41}. Cancel out k^{52} in both numerator and denominator.

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

Differentiate w.r.t. k

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

Steps Using Chain Rule

To multiply powers of the same base, add their exponents. Add 80 and -28 to get 52.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{k^{52}}{k^{93}})$$

Rewrite k^{93} as k^{52}k^{41}. Cancel out k^{52} in both numerator and denominator.

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

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).

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

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}.

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

Simplify.

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