Find dxd(n+1xn+1+c) and hence deduce that ∫xn dx=n+1xn+1+c.
From Table 10.1 we find
dxd(n+1xn+1+c)=dxd(n+1xn+1)+dxd(c) using the linearity of differentiation
=n+11dxd(xn+1)+dxd(c) again using the linearity of differentiation
=n+11{(n+1)xn}+0 using Table 10.1=xn
Consequently, reversing the process we find
∫xn dx=n+1xn+1+c
as required. Note that this result is invalid if n = −1 and so this result could not be applied to the integral ∫(1/x)dx.
Table 10.1 Derivatives of commonly used functions. |
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Function, y(x) | Derivative, y′ | Function, y(x) | Derivative, y′ |
constant | 0 | cos−1(ax+b) | 1−(ax+b)2−a |
xn | nxn−1 | ||
ex | ex | tan−1(ax+b) | 1+(ax+b)2a |
e−x | −e−x | ||
eax | aeax | sinh(ax+b) | acosh(ax+b) |
cosh(ax+b) | asinh(ax+b) | ||
lnx | x1 | tanh(ax+b) | asech2(ax+b) |
sinx | cosx | cosech(ax+b) | −acosech(ax+b)× |
cosx | −sinx | coth(ax+b) | |
sin(ax+b) | acos(ax+b) | sech(ax+b) | −asech(ax+b)× |
cos(ax+b) | −asin(ax+b) | tanh(ax+b) | |
tan(ax+b) | asec2(ax+b) | coth(ax+b) | −acosech2(ax+b) |
cosec(ax+b) | −acosec(ax+b)cot(ax+b) | sinh−1(ax+b) | (ax+b)2+1a |
sec(ax+b) | asec(ax+b)tan(ax+b) | (ax+b)2−1a | |
cot(ax+b) | −acosec2(ax+b) | cosh−1(ax+b) | (ax+b)2−1a |
sin−1(ax+b) | 1−(ax+b)2a | tanh−1(ax+b) | 1−(ax+b)2a |