Solve b^-4b^3

$$b ^ { - 4 } b ^ { 3 }$$

Differentiate w.r.t. b

$$-\frac{1}{b^{2}}$$

Steps Using Derivative Rule for Product

For any two differentiable functions, the derivative of the product of two functions is the first function times the derivative of the second plus the second function times the derivative of the first.

$$b^{-4}\frac{\mathrm{d}}{\mathrm{d}b}(b^{3})+b^{3}\frac{\mathrm{d}}{\mathrm{d}b}(b^{-4})$$

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}.

$$b^{-4}\times 3b^{3-1}+b^{3}\left(-4\right)b^{-4-1}$$

Simplify.

$$b^{-4}\times 3b^{2}+b^{3}\left(-4\right)b^{-5}$$

To multiply powers of the same base, add their exponents.

$$3b^{-4+2}-4b^{3-5}$$

Simplify.

$$3b^{-2}-4b^{-2}$$

Evaluate

$$\frac{1}{b}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add -4 and 3 to get -1.

$$b^{-1}$$