Solve b^85/b^31*b^90

$$\frac { b ^ { 85 } } { b ^ { 31 } \cdot b ^ { 90 } }$$

Evaluate

$$\frac{1}{b^{36}}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 31 and 90 to get 121.

$$\frac{b^{85}}{b^{121}}$$

Rewrite b^{121} as b^{85}b^{36}. Cancel out b^{85} in both numerator and denominator.

$$\frac{1}{b^{36}}$$

Differentiate w.r.t. b

$$-\frac{36}{b^{37}}$$

Steps Using Chain Rule

To multiply powers of the same base, add their exponents. Add 31 and 90 to get 121.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{b^{85}}{b^{121}})$$

Rewrite b^{121} as b^{85}b^{36}. Cancel out b^{85} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{1}{b^{36}})$$

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).

$$-\left(b^{36}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}b}(b^{36})$$

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}.

$$-\left(b^{36}\right)^{-2}\times 36b^{36-1}$$

Simplify.

$$-36b^{35}\left(b^{36}\right)^{-2}$$