Solve 7/6b^3*(-30/49b)=

$$\frac { 7 } { 6 } b ^ { 3 } \cdot ( - \frac { 30 } { 49 } b ) =$$

Evaluate

$$-\frac{5b^{4}}{7}$$

Solution Steps

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

$$\frac{7}{6}b^{4}\left(-\frac{30}{49}\right)$$

Multiply \frac{7}{6} and -\frac{30}{49} to get -\frac{5}{7}.

$$-\frac{5}{7}b^{4}$$

Differentiate w.r.t. b

$$-\frac{20b^{3}}{7}$$

Steps Using Definition of a Derivative

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

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{7}{6}b^{4}\left(-\frac{30}{49}\right))$$

Multiply \frac{7}{6} and -\frac{30}{49} to get -\frac{5}{7}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(-\frac{5}{7}b^{4})$$

The derivative of ax^{n} is nax^{n-1}.

$$4\left(-\frac{5}{7}\right)b^{4-1}$$

Multiply 4 times -\frac{5}{7}.

$$-\frac{20}{7}b^{4-1}$$

Subtract 1 from 4.

$$-\frac{20}{7}b^{3}$$