Solve =3x*16x^6y^4-8x^6y^3*5xy

$$= 3 x \cdot 16 x ^ { 6 } y ^ { 4 } - 8 x ^ { 6 } y ^ { 3 } \cdot 5 x y$$

Evaluate

$$8y^{4}x^{7}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 6 to get 7.

$$3x^{7}\times 16y^{4}-8x^{6}y^{3}\times 5xy$$

To multiply powers of the same base, add their exponents. Add 6 and 1 to get 7.

$$3x^{7}\times 16y^{4}-8x^{7}y^{3}\times 5y$$

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

$$3x^{7}\times 16y^{4}-8x^{7}y^{4}\times 5$$

Multiply 3 and 16 to get 48.

$$48x^{7}y^{4}-8x^{7}y^{4}\times 5$$

Multiply 8 and 5 to get 40.

$$48x^{7}y^{4}-40x^{7}y^{4}$$

Combine 48x^{7}y^{4} and -40x^{7}y^{4} to get 8x^{7}y^{4}.

$$8x^{7}y^{4}$$

Differentiate w.r.t. x

$$56y^{4}x^{6}$$

Steps Using Derivative Rule for Sum

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}.

$$48y^{4}x^{6}x^{1-1}+6\left(-40xy^{4}\right)x^{6-1}$$

Subtract 1 from 1.

$$48y^{4}x^{6}x^{0}+6\left(-40xy^{4}\right)x^{6-1}$$

Multiply 6 times -8\times 5y^{3}xy.

$$48y^{4}x^{6}x^{0}+\left(-240xy^{4}\right)x^{6-1}$$

Subtract 1 from 6.

$$48y^{4}x^{6}x^{0}+\left(-240xy^{4}\right)x^{5}$$

For any term t except 0, t^{0}=1.

$$48y^{4}x^{6}\times 1+\left(-240xy^{4}\right)x^{5}$$

For any term t, t\times 1=t and 1t=t.

$$48y^{4}x^{6}+\left(-240xy^{4}\right)x^{5}$$