Derivative Rules for y=cos(x) and y=tan(x)
Key Questions
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The derivative of
tanxtanx issec^2xsec2x .To see why, you'll need to know a few results. First, you need to know that the derivative of
sinxsinx iscosxcosx . Here's a proof of that result from first principles:Once you know this, it also implies that the derivative of
cosxcosx is-sinx−sinx (which you'll also need later). You need to know one more thing, which is the Quotient Rule for differentiation:Once all those pieces are in place, the differentiation goes as follows:
d/dx tanxddxtanx
=d/dx sinx/cosx=ddxsinxcosx =(cosx . cosx-sinx.(-sinx))/(cos^2x)=cosx.cosx−sinx.(−sinx)cos2x (using Quotient Rule)=(cos^2x+sin^2x)/(cos^2x)=cos2x+sin2xcos2x =1/(cos^2x)=1cos2x (using the Pythagorean Identity)=sec^2x=sec2x -
Using the definition of a derivative:
dy/dx = lim_(h->0) (f(x+h)-f(x))/hdydx=limh→0f(x+h)−f(x)h , whereh = deltaxh=δx We substitute in our function to get:
lim_(h->0) (cos(x+h)-cos(x))/hlimh→0cos(x+h)−cos(x)h Using the Trig identity:
cos(a+b) = cosacosb - sinasinbcos(a+b)=cosacosb−sinasinb ,we get:
lim_(h->0) ((cosxcos h - sinxsin h)-cosx)/hlimh→0(cosxcosh−sinxsinh)−cosxh Factoring out the
cosxcosx term, we get:lim_(h->0) (cosx(cos h-1) - sinxsin h)/hlimh→0cosx(cosh−1)−sinxsinhh This can be split into 2 fractions:
lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/hlimh→0cosx(cosh−1)h−sinxsinhh Now comes the more difficult part: recognizing known formulas.
The 2 which will be useful here are:
lim_(x->0) sinx/x = 1limx→0sinxx=1 , andlim_(x->0) (cosx-1)/x = 0limx→0cosx−1x=0 Since those identities rely on the variable inside the functions being the same as the one used in the
limlim portion, we can only use these identities on terms usinghh , since that's what ourlimlim uses. To work these into our equation, we first need to split our function up a bit more:lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/hlimh→0cosx(cosh−1)h−sinxsinhh becomes:
lim_(h->0)cosx((cos h-1)/h) - sinx((sin h)/h)limh→0cosx(cosh−1h)−sinx(sinhh) Using the previously recognized formulas, we now have:
lim_(h->0) cosx(0) - sinx(1)limh→0cosx(0)−sinx(1) which equals:
lim_(h->0) (-sinx)limh→0(−sinx) Since there are no more
hh variables, we can just drop thelim_(h->0)limh→0 , giving us a final answer of:-sinx−sinx . -
dy/dx=(cos x)'=-sin x . See derivatives of trig functions for details.
Questions
Differentiating Trigonometric Functions
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Limits Involving Trigonometric Functions
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Intuitive Approach to the derivative of y=sin(x)
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Derivative Rules for y=cos(x) and y=tan(x)
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Differentiating sin(x) from First Principles
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Special Limits Involving sin(x), x, and tan(x)
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Graphical Relationship Between sin(x), x, and tan(x), using Radian Measure
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Derivatives of y=sec(x), y=cot(x), y= csc(x)
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Differentiating Inverse Trigonometric Functions