Tan X + Sec X
Derivative of sec x
Sec x is another way of writing 1/cos x. Now nosotros need to apply the quotient dominion. The quotient rule is that y dash equals (vu dash – uv nuance) / v^2 divided by v square. In this case, we accept y equals to sec x. Yet, we are going to let u equal the numerator of 1 and v equal the denominator: cos x. So after applying the caliber dominion, we have v, cos remains unchanged and multiply past u.
So the derivative of 1, the constant, is goose egg. And then minus u, which is one, times the derivative, and that is the derivative of cos 10. Now that is equal to minus sin x all over v^2, which is cos^2 x. Then the consequence is positive sin x divided by the cos ^two 10. Now rewrite this as [sin x/cos ten] times 1/cos ten. Still, the sin x divided by cos square x is equal to the tangent of x. Also, the ane/cos ten is equal to the secant of x. So. the derivative of sec x is equal to the tan 10 multiplied past sec x.
And so, nosotros take f(x) = 1/cos x = u/v
By caliber rule,
f'(10) = (vu' – uv') / v2
Now, f'(10) = [cos x d/dx(1) – 1 d/dx(cos ten)] / (cos x)2
= [cos x (0) – one (-sin x)] / cos2x
= (sin x) / cos2x
= 1/cos 10 · (sin x)/(cos ten)
= sec x · tan 10
Derivative of sec x tan ten
We have the derivative of the secant of ten times the tangent of x plus the secant of x times the derivative of the tangent of x. Next, we need to utilise the product dominion, and the dominion is f times the g. The f part is the secant x, and g is equal to tangent 10. Now, f dash is equal to secant 10 times tangent ten. Too, 1000 nuance is equal to secant square x. Now we are going to pair things up, y dash is equal to secant 10 times secant squared ten plus tangent x times secant x tangent x. Next, we need to simplify this expression. Now we put the secant x to the front end. Then secant 10 times secant squared x plus tangent foursquare x. This is the outcome of sec ten tan x.
y=sec x tan x
Past product dominion, f= secx, grand=sec^2x and f dash= secx tanx, g nuance= sec^2x
Adjacent, y dash= sec x⋅sec^2 ten + tan ten⋅sec ten tan x
Side by side, y nuance= sec x (sec^2 ten + tan^two x)
Derivative of sec square x
Let, sec x is equal to y. At present the derivative of sec squared x equals the derivative y square. So the derivative of y square's upshot is the derivative of 2y. Now, nosotros have y equals sec x. After applying this, we get 2 sec x times derivative of sec x, which is equal to 2sec x times sec 10 tan x. At last, the derivative of sec square x is 2 sec squared x times tan x.
Permit, y = sec x.
By ability rule and chain rule, y'(x) = 2 secx d/dx (sec x)
Next, y'(x)= 2 sec x · (sec x · tan ten)
Side by side, y'(x)= ii sec2x tan x
Derivative of sec x degree
We have the derivative of sec x degree. So, y is equal to sec 10 caste. However, nosotros know the pi value is 180 degrees. Then 10 degree equals (π/180)10 radians. So, y equals sec (π/180) 10. At present the derivative of y is equal to the sec of (π/180) times tan of (π/180) times (π/180). And then the derivative of sec x degree is (π/180) sec x degree times tan x degree.
y = sec 10 degree
We know π radian = 180 degree
Ten caste = (π/180)10 radians
Then y = sec (π/180)x
dy/dx = sec (π/180)x tan (π/180)10 (π/180)
= (π/180) sec xo tan x0
Derivative of tan^2 x
We have the derivative of tan square x. And then, let y be equal to tan square ten. Differentiate with respect to 10, dy upon dx equals the derivative of tan square x. Now it will be tan x whole square upon d tan x into d tan 10 upon dx. After we apply the xn formula, information technology will be two tan ten times sec squared ten. Then the derivative of tan square ten is 2 tan x sec squared x.
Let, y = tan^2x
dy/dx = d/dx (tan^2x) = d(tanx)^2 /d(tanx) times d(tanx)/dx
Next, dy/dx= 2 tanx times sec^2x
Next = dy/dx = 2tan sec^2x.
Derivative of log (sec 10 + tan x)
We have a derivative of the log of sec 10 plus tan x. And so, let y be equal to the log of sec x plus tan x. Differentiate with respect to 10, dy upon dx equal to the derivative of log sec x plus tan x. Nosotros tin can't use the direct formula of log in this equation. So, we use log sec ten plus tan 10 upon sec x plus tan x into the derivative of sec x plus tan x upon dx. However, log x is equal to ane, and log sec 10 plus tan ten equals sec x times tan ten plus sec squared x. And then nosotros common the sec x, and information technology volition be (sec x+ tan x) upon (sec 10+ tan x). Next, both (sec x+ tan x) are canceled. At last, the derivative of log sec ten plus tan x is equal to sec x.
Let y = (log (sec x + tan x))
Differentiate w.r.t.x
dy/dx= 1/(sec x + tan 10) × d/dx (sec x + tan x)
Next, dy/dx = [1/(sec x + tan ten) ]× sec 10 tan x + sec2x
= [1/(sec x + tan ten)] × sec x (tan 10 + sec x)
= sec x
Derivative of y=tan x -1/ sec 10
We take a derivative of y equals tan x minus one upon sec ten. And so, y dash tin can be equal to the tangent of an x minus one prime secant of x minus tan x minus one time the secant of x prime. Next, we get the derivative of the tangent, which is secant squared ten and times secant of x, minus (tangent of x minus 1) times sec x tan x. So we will get sec cube x minus tan squared x sec x plus sec x tan x. After some calculation, we have an identity which is sec squared x minus tan squared 10 plus tan x upon sec x. At terminal, the derivative of tan x -1/ sec x is 1 plus tan x upon sec x. Read Also:Derivative of Sin2x | Formula, Proof, Rules & Examples
Y dash = (tanx-1)dash sec 10 – (tanx-1) sec x dash = sec^2x · sec ten- (tanx-1)sec x tan 10
Y dash = sec^3x – tan^2x sec x + secx tanx
Next, y dash = sec^2x-tan^2x +tanx/sec x = one +tan x/sec x
Derivative of tan x^ cot x
- Let y equals tan x to the power cot x. The showtime method, a to the power b equals to e to the power b log a. And so, y equals to due east to the power cot x log tan x. Differentiate with respect to 10, dy upon dx equals to e to the power cot x log tan x into differentiate cot x into log tan 10. Now, Nosotros will use the product dominion. And then we get tan x to the ability cot x times cot x into the differentiation of log tan 10 plus log tan x into the differentiation of cot x.
- Now, it equals tan x to the ability cot x into cot x times 1/tan ten into sec square 10 plus log tan x into minus cos foursquare 10. Adjacent, it will exist tan x to the power cot ten, and in the tertiary brackets, cos ten/sinx into cos x/sin x into 1/cos^2x minus log tan x into cosec^2x. Now we are canceling the cos x. So dy/dx equals tan x to the ability cot x into cosec square x minus cosec foursquare ten into log tan x. After some calculation, the answer is tan to the power cot x into cosec^2x into one minus log tan x.
Permit, y=tanx^cot(10)
Then, log y = log tan(10)^cot(x)
Then, log y= cot(ten).log(tanx).
Side by side, i/y.dy/dx=cotx.(ane/tanx).sec^2x-cosec^2x.log tanx.
Adjacent, 1/y.dy/dx=cosec^2x-cosec^2x.log tan x
Now, dy/dx=y.cosec^2x(1-log tanx)
Now, dy/dx=tanx^cotx.cosec^2x.(1-log tan x)
Graph of sec x
At first, the numbers are going to intersect at ane minus one and back up at 1 again. Next we have asymptotes and ninety degrees, 270 degrees, because we cant take ane over 0. And then the graph is going to fit around doing the opposite of decimal. And then, 1 over this decimal betwixt 0 and 90 is gonna show u.s.a. a big value. 1 over the negative decimals between ninety and 270 are going to requite us bigger negative values. Side by side one over values of the decimal were between 270 and 360 are gonna give united states large values. However, intersection points at 0 degrees, 360 degrees and 180 degrees are undefined whenever the graph is equal to zero.
Integral of sec x dx
- If we write the secant of x as one over the cosine of ten which is its definition, we endeavor to integrate; unfortunately this expression here is not integratable. Nevertheless nosotros are going to take to manipulate this integrand to get into a form that we can integrate. Now in that location are probably a multitude of means nosotros tin can exercise this. But we can practice this with two mutual ones. Starting time one is to employ partial fractions. Now in this first method, it will accept quite a few steps and information technology is a robust method. Because in the second method, it requires a flake more intuition and information technology requires converting this integral into the form of f prime ten over f of x. And so the result is going to be the log of the part. And so, basically the second method uses a u substitution.
- And then, in the partial fractions, we have established sec x that can be written equally 1 over the cos of x. At present, this is equivalent if we multiply the top and lesser of this fraction past cos of 10. So, nosotros are multiplying by one, the height becomes cos of x and the lesser becomes cos squared of x. Side by side, we tin can rewrite this considering the identity (cosine squared 10 + sine squared 10) is equal to one. Then we can rewrite the bottom or the denominator equally (1 – sine squared 10).
- Next we can rewrite the sec ten every bit sec x multiplied by sec x plus tan 10 divided by sec x plus tan 10. Then we demand to multiply this by i and we expand sec x into the numerator. However, we get sec squared x plus sec x tan x over the denominator (sec ten+ tan 10). Now, the integral of sec x is completed.
Formula of Sec x
The formula of secant of x is 1 divided by the cosine of x: sec 10 = 1 cos x.
Some frequently asked questions
What is the derivative of tan sec x?
Sec 2x is the derivative of tan sec x.
Is SEC 2x always positive?
Yes, sec 2x is always positive.
What is tan 2x equal to?
Sin 2x/cos 2x is equal to tan 2x.
Is tan 2x 1 Sec x a Pythagorean identity?
Yes, tan 2x 1sec x is a Pythagorean identity.
What is the relation betwixt Tan X and Sec x?
The relation between tan 10 and sec 10 are tan x = sin 10 cos ten.
What is tan 1x?
tan−1x = tan−1(x).
What is the derivative of SEC 2x?
f'(x) = two secx d/dx(sec x) = 2 sec x (sec x tan ten) = 2 sec2x tan x is the derivative of sec 2x.
Tan X + Sec X,
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