## The Calculus Primer (2011)

### Part III. Differentiation of Algebraic Functions

### Chapter 9. DERIVATIVE OF THE POWER FUNCTION

**3—5. The Derivative of a Variable with a Constant Exponent.** Consider the power function *y* = *v ^{n}*. Applying the General Rule, and expanding by the Binomial Theorem, we obtain:

*Step* 1.*y* + Δ*y* = (*v* + Δ*v*)* ^{n}*.

*Step* 2.Δ*y* = (*v* + Δ*v*)* ^{n}* −

*v*.

^{n}*Step* 3.

Now, as Δ*x* → 0, Δ*v* → 0; hence

*Step* 4.

RULE. *The derivative of a variable with a constant exponent is equal to the product of that exponent, the variable raised to a power one less than the original exponent, and the derivative of the variable*.

NOTE 1.It is assumed, in the above proof, that the exponent *n* is a positive integer. If this were not so, we could not have applied the binomial theorem, for when *n* is negative or fractional, the number of terms in the binomial expansion is infinite; we could not pass to the limit, therefore, since Principle (III) of §1—15 applies only to a finite number of terms. However, we shall find later that the relation

holds for any rational value of *n*, whether positive, negative, or fractional.

NOTE 2.As a special case of formula [5] above, we let *v* = *x*, and obtain:

NOTE 3.Formulas [5] and [5a], in view of what was said in NOTE 1, may be applied to a power function when the exponent is negative or fractional, as shown in Examples 3 and 4 below; thus

EXAMPLE 1.Differentiate *y* = *x*^{3} + 5*x*^{2} + 4.

*Solution*. = 3*x*^{2} + 10*x*.

EXAMPLE 2.Differentiate *y* = 5*x*^{4} + 6*x*^{3} − 3*x*^{2} + 9*x* − 21.

*Solution*. = 20*x*^{3} + 18*x*^{2} − 6*x* + 9.

EXAMPLE 3.Differentiate *y* =

*Solution*.*y* = *x*^{−3} + 3*x*^{−2} − 5*x*^{−1}.

= −3(*x*^{−4}) + (3)(−2)(*x*^{−3}) − 5(−1)(*x*^{−2}),

or

EXAMPLE 4.Differentiate

*Solution*. *y* = 2*x*^{½} + (3*x*)^{½} − 4(*x*^{−⅓}).

EXAMPLE 5.Find :*y* = (5*x* + 3)^{2}.

*Solution*.*v* = 5*x* + 3, and *y* = *v*^{2}.

Hence = *nv*^{n−1} = 2*v*(5) = 10(5*x* + 3).

This result may be verified by expanding the given function before differentiating; thus

*y* = (5*x* + 3)^{2} = 25*x*^{2} + 30*x* + 9;

= 50*x* + 30 = 10(5*x* + 3).

EXAMPLE 6.Differentiate with respect to *x*: *y* = 4 (*x*^{2} + 6*x*)^{3}.

*Solution*.

Here *v* = *x*^{2} + 6*x*, and *y* = 4*v*^{3}.

Hence = *nv*^{n−1} = (4)(3)*v*^{2}

= (4)(3)(*x*^{2} + 6*x*)^{2}(2*x* + 6)

= 12(*x*^{2} + 6*x*)^{2}(2*x* + 6).

EXAMPLE 7.Differentiate *y* =

*Solution*.*y* = (*x*^{2} − 4)^{½};*v* = *x*^{2} − 4

**EXERCISE 3—1**

*Differentiate:*

**1.***y* = 2*x*^{3} − 5*x* + 8

**2.***y* = *x*^{5} − 3*x*^{4} + 2*x*^{3} + *x*

**3.***y* =

**4.***y* = 5 + 6

**5.***y* = *v*_{1}*t* + *gt*^{2}

**6.***y* = *x*^{−3} − 3*x*^{−2} + 4*x*^{−1}

**7.***y* =

**8.***s* =

**9.***y* = (*x*^{2} − 5*x*)^{2}

**10.***s* = *k*(*a* + 2*t*)^{3}

**11.***y* = *x*^{m} − *mx*^{3}

**12.***y* =

**13.***y* = (*x*^{3} − 2*x*)^{4}

**14.***y* =

**15.***y* =

**16.***y* =