k -fx h x, 133 are but different same thing. Observe further that y', being equal to a 3x2, is also found; so that from we have a function of x and its derivative may be y' = f'x = = a • 3x2, which is the derivative of y', or the derivative of the derivative of y, or simply the second derivative of y. In like manner the third derivative of y would be indicated by y'" or f""x, the fourth by yi or fir, &c. PROPOSITION 1. To find the derivative of any function capable of being developed in integral additive powers of the variable. Let the function be denoted by ... .... (242) y = ƒx = α2+a1x1 + а‚x2 + аžÃ3 + +a,xn+.. Increasing x by h and the function by k, we have y + k = f(x + h) = a + a1(x + h)1 + a2(x+h)2 + ... + a‚(x+h)”+ ..., .. k=f(x+h) — ƒx = a ̧[(x + h)1 − x1] + α2[(x + h)2 − x2]+... +a„[(x+ h)” — x”]+....., -- but h = (x+h)2x2 h where X is such a function of x and h that Xh shall disappear when But to prove this let us put x+h=a, x=b, and ... ha−b; h is diminished to zero. so that we fall upon the examples under (16), and we have only to solve the problem whether a”- b" is divisible by a- - b. In order to this we execute the multiplication, (a - b) (a"1+a"-2b+an-3 b2 + amb3 + am-5 b1 + ... + ab2 + b2), and we find the product to be -1 3 a”+ an1b+an—2 b2 + an3 b3 + am b1 + ... + a2b”—2 + ab”—1 Hence, the difference of the same powers of any two quantities is divisible by the difference of the quantities themselves, and the quotient is homogeneous and one degree lower than the power, the leading quantity descending one degree each term and the following ascending by the same law. If in (243) we now replace a and b by their values x+h and x, we find (x + h)" - x" h ·= (x + h)"~2 +(x+h)"2x + (x+h)"3x2+...+x-1; (x+h)”-2x+(x+h)”—3x2 whence the fourth equation becomes k = h f(x+h)-fx · = a1 + a12 (2x+h) + a3 (3x2 +3xh+h2) + ... +an [(x+h)+(x+ h)" 2x + (x + h)" 3 x2+...+x-1]; therefore, making h = 0 and observing that there are ʼn terms in the expression (x+h)-1+..., which all reduce to ", we get y' = f'x = [ * ] = [( f ( x + h) ̄f(x+h) —ƒx ̄ = a1+ a2 • 2x+az • 3x2 (244) +α1• 4x3 + +a, • nx11; which is the derivative sought. ... We perceive that (244) is derived from (242) by multiplying by the exponents of x in the several terms and decreasing them by DERIVATIVE OF Ar". 135 unity; and hence that the term a, independent of x, or, which is the same thing, the term a。 • xo, disappears in (244); so that, if we were to pass from (244) back to (242), it would be necessary to introduce a term independent of x, or to add a constant quantity, which may be represented by C. PROPOSITION II. To find the Derivative of any real power of a variable. Let y = fx = Ax" be the function, a being any whatever, whole or fractional, plus or minus, but not imaginary, not of the form c. Suppose, in the first place, that a is fractional and @= 2n Suppose then that, while x is increased by h, z receives the increment i, and, as a consequence, that y is augmented by k; we have y + k = A(z+i)”, and z + i = (x + h)*, or (z+i)" = x+h; k = A(z+i)" - Az", and (z+i)" — z" = h ; .'. (z+i)m — zm *y'y=f, read, the derivative of y, y being a function of x. =A. • xn n m 1 m m A • xn n Here it is worthy of remark that we have found the derivative of y a function of x, by taking the product of the derivative of y a function of z, and that of z a function of x. We observe the same rule, then, for the derivative, whether the exponent be fractional or integral, we multiply by the exponent, and diminish it by unity. Let a be now taken subtractive, and either integral or fractional, or let a = −r, where r is either a whole number or a fraction; we have To find the Derivative of a variable, affected by any (245) exponent whatever that is not imaginary; multiply the variable by its exponent, and decrease that exponent by unity. Thus, the derivative of x2 is 2x2-1, of x3 is 3x2, of x is 4x3; of ; of Ax" is Aax"; of x is 1 • x11 = 1 •.x° 1; of A, (= Axo), it is A.0.x-1 =A.0. = 0. DEVELOPED FUNCTIONS. 137 PROPOSITION III. The Derivative of any function, capable of being devel- (246) oped in real powers of the variable, will be found by multiplying in each term by the exponent, and decreasing it one. CONVERSELY if a derived function be developed in powers of the variable, we shall return to the original function, by increasing the exponents by unity, and dividing by the exponents thus augmented, taking care to add a constant. Thus, let y be such a function of x, or depend upon z in such way, that we have y = fx = Axa + BÃ3 + Сx2 + ..., (247) where A, a; B, b; C, c; &c., may be any real quantities, + or whole or fractional, we have (245) y' = f'x = .... (248) We observe that if a = = 0, or there be a constant term, Ax0 = A, in (247), it will disappear in (248), since Aar then becomes A.0 • 1 =0; therefore, in passing back from (248) to (247) we must add a constant for that which may have disappeared. We shall see how this constant will be determined in any particular case by the nature of the problem. To illustrate, suppose we have found the derivative y' = 3x — fxt +7, returning to the function (246) we obtain y = ‡x2 — f • 4x2+7x+ constant. Whenever, then, we can find the derived function developed in powers of the variable, there will be no difficulty in ascertaining the primitive function. PROPOSITION IV. The Derivatives of equal functions, depending upon (249) the same variable, are themselves equal. This is manifest from the nature of the operation; thus, if we have any two functions of x, such that Fx = fx, |