Solve (b^-7)^2/(b^3)^3=

$$\frac { ( b ^ { - 7 } ) ^ { 2 } } { ( b ^ { 3 } ) ^ { 3 } } =$$

Evaluate

$$\frac{1}{b^{23}}$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply -7 and 2 to get -14.

$$\frac{b^{-14}}{\left(b^{3}\right)^{3}}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$\frac{b^{-14}}{b^{9}}$$

Rewrite b^{9} as b^{-14}b^{23}. Cancel out b^{-14} in both numerator and denominator.

$$\frac{1}{b^{23}}$$

Differentiate w.r.t. b

$$-\frac{23}{b^{24}}$$

Steps Using Chain Rule

To raise a power to another power, multiply the exponents. Multiply -7 and 2 to get -14.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{b^{-14}}{\left(b^{3}\right)^{3}})$$

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{b^{-14}}{b^{9}})$$

Rewrite b^{9} as b^{-14}b^{23}. Cancel out b^{-14} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{1}{b^{23}})$$

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).

$$-\left(b^{23}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}b}(b^{23})$$

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}.

$$-\left(b^{23}\right)^{-2}\times 23b^{23-1}$$

Simplify.

$$-23b^{22}\left(b^{23}\right)^{-2}$$