Techniques for differentiation

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This article contains a list of techniques for the differentiation of real functions, categorized by type.

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[edit] Simple polynomial functions

Given a polynomial p(x), that is defined by the formula:

p(x) = \sum^m_{i=0} k_i x^i , one has
\frac{d}{dx} p(x) = \sum^m_{i=0} ik_ix^{i-1}.

That is, one simply multiplied each term by its degree, then divides by ‘’x’’. For example, one can differentiate \sqrt{x} + 5x . First, one would break it down into its component terms: sqrt(x) and 5x. sqrt(x) is equal to x1/2, meaning that its derivative is 1/(2sqrt(x)), or half the reciprocal of the value. 5x simply becomes 5, giving us:

\frac{d}{dx} (\sqrt{x} + 5x) = \frac{1 + 10\sqrt{x}}{2\sqrt{x}}.

[edit] Exponential functions

Given some function ‘’f(x)’’ equals bx, its derivative can be found via the following formula:

\frac{d}{dx} b^x = b^x \ln b

where ‘’ln b’’ is the natural logarithm of b. Using this formula, we can differentiate 225x, which gives us 2(ln 3 + ln 5). (See Natural logarithm). So ultimately, we have bx 2ln 3 + bx 2ln 5.

[edit] Proof

\frac{d}{dx}b^x
=\frac{d}{dx}e^{\ln b^x} A property of logarithms.
=\frac{d}{dx}e^{x\ln b} Another property of logarithms
=\left(\frac{d}{dx}x\ln b\right)\left(e^{x\ln b}\right) From the chain rule.
=\left(\ln b\right)\left(e^{\ln b^x}\right)
= bxlnb

[edit] Logarithmic functions

All logarithmic functions can be differentiated via a formula very similar to that for exponential functions. The slope of any logarithmic function at a point x is equal to the reciprocal of x times the natural logarithm of the base, or:

\frac{d}{dx} \log_b x = \frac{1}{x \ln b}.

Through this we can differentiate the natural logarithm itself. Of course, the base of the natural logarithm is e, and the base x logarithm of x is always one. Therefore, the natural logarithm of e is one. Knowing this, we can find that the slope of the natural logarithm at any point equals the reciprocal of the height at that point.

[edit] Proof

Let y = logbx.

Then by = x.

e^{\ln b^y}=x
eylnb = x

Use implicit differentiation.

\left(e^{y\ln b}\right)\left(\frac{d}{dx}y\ln b\right)=1
\left(e^{\ln b^y}\right)\left(\frac{dy}{dx}\right)\left(\ln b\right)=1
\frac{dy}{dx}=\frac{1}{(b^y)(\ln b)}

Since y = logbx and by = x, \frac{d}{dx}\log_b x=\frac{1}{x\ln b}.

[edit] Simple Trigonometric functions

\frac{d}{dx}\sin x=\cos x

\frac{d}{dx}\cos x=-\sin x

\frac{d}{dx}\tan x=\sec^2 x

\frac{d}{dx}\sec x=\tan x \sec x

\frac{d}{dx}\csc x=-\cot x \csc x

\frac{d}{dx}\cot x=-\csc^2 x

For an extensive list of derivatives of trigonometric functions, hyperbolic functions, their inverses, and proofs, see table of derivatives and Differentiation of trigonometric functions.

[edit] See also