Solve j^-29/j^-7*j^-9

$$\frac { j ^ { - 29 } } { j ^ { - 7 } \cdot j ^ { - 9 } }$$

Evaluate

$$\frac{1}{j^{13}}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add -7 and -9 to get -16.

$$\frac{j^{-29}}{j^{-16}}$$

Rewrite j^{-16} as j^{-29}j^{13}. Cancel out j^{-29} in both numerator and denominator.

$$\frac{1}{j^{13}}$$

Differentiate w.r.t. j

$$-\frac{13}{j^{14}}$$

Steps Using Chain Rule

To multiply powers of the same base, add their exponents. Add -7 and -9 to get -16.

$$\frac{\mathrm{d}}{\mathrm{d}j}(\frac{j^{-29}}{j^{-16}})$$

Rewrite j^{-16} as j^{-29}j^{13}. Cancel out j^{-29} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}j}(\frac{1}{j^{13}})$$

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).

$$-\left(j^{13}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}j}(j^{13})$$

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}.

$$-\left(j^{13}\right)^{-2}\times 13j^{13-1}$$

Simplify.

$$-13j^{12}\left(j^{13}\right)^{-2}$$