# Derivative of Tanx

By | July 10, 2020

# Derivative of Tanx

Derivative of Tanx,I want to differentiate derivative of tanx. How am I going to do this? Well, hopefully you remember that tangent theta is sine theta over cosine theta. Why is this? If you think back to the business about the right triangles, right, sine is this opposite side that I’m calling y over the hypotenuse.

Derivative of Tanx

And cosine is this adjacent side whose length I’ll call x over the hypotenuse. I’ll call this angle theta. Then sine theta is y over r, and cosine theta is x over r, so this fraction can be simplified to y over x. That’s the opposite side over the adjacent side. That’s tangent of theta.

## Derivative of Tanx

So good. I can express Tangent theta as sin theta over cosine theta, how does that help? Well, I know how to differentiate sine and cosine, and by writing tangent this way, tangents now a quotient, so I can use the. Quotient rule. All right, so I’ll use the quotient rule, and the derivative of tangent theta is the derivative of the numerator, which is cosine, times the der, denominator, which is just cosine, minus the derivative of the denominator, which is minus sine, times the numerator, which is sine.

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Derivative of Tanx

And this whole thing is over the denominator squared, cosine squared theta. Now, is this very helpful? Well, I’ve got cosine times cosine, so I can write that as cosine squared theta. And here I’ve got minus negative sine theta times sine theta, so that ends up being plus sine squared theta. And the whole thing is still being divided by cosine squared theta. (Derivative of Tanx) Can I simplify that at all? Well cosine squared plus sine squared, that’s the Pythagorean identity. That’s just one. So I can replace the numerator with just one, still over cosine squared theta. And one over cosine squared theta, well, one over cosine is called secant, so this is really secant squared theta. . So what all this shows is the derivative of tangent theta is second squared theta.

A moment ago we did a calculation using the quotient rule to see that the derivative of tangent theta is second squared theta. And now I want to see how this plays out in some concrete example to get a, a real sense as to why a formula like this is true. So here’s a couple triangles. They’re both right triangles and the length of their hypotenuse is the same. This angle I am calling alpha, let’s this be a (Derivative of Tanx) little bit less than 45 degree. And this angle I am calling beta, and it’s more than 45 degrees. Now what you can say about the Secant of alpha and the Secant of beta? You know, the Secant of alpha is definitely bigger than one. I mean, what’s the Secant? It’s this hypotenuse length divided by this length here. And that’s bigger than one. How does it compare to beta? Well, the secant of beta is quite a bit bigger than the secant of alpha, and why is that? Well,

the secant of beta is this length here, the hypotenuse, divided by this width, but this triangle is quite a bit narrower than this triangle. So the secant has the same numerator, the hypotenuse, the same length, but the denominator here is quite a bit smaller, and if the denominator’s a lot smaller, then the ratio, which is the secant, is quite a bit bigger. Some of the secan of beta is bigger than the secan of alpha, and the secan of alpha is bigger than one. And that means that secan squared alpha is also (Derivative of Tanx) bigger than one. And secan squared alpha is less than secan squared beta. And the significance of that is right here, the derivative of tangent theta is secan squared theta. So what does this mean? Well, this, this is telling me how wiggling theta affects tangent. It affects it like a factor of secant squared theta. So in this example, where secant squared beta is a lot bigger than secant squared alpha, the effect of wiggling beta on the tangent of beta should be a lot larger than the effect of wiggling alpha on the tangent of alpha. And you can see that. Let me draw a triangle, where I’ve wiggled the angle alpha up a little bit. I’ve made it a little bit bigger. But I’m going to make the same hypotenuse.

All right, so this hypotenuse length is the same as this length, but I’ve made the angle a little bit bigger. And how is the tangent of the slightly larger alpha compared to the tangent of alpha. Well, the tangent of the slightly larger alpha is bigger than tangent alpha, but not by all that much. Now compare that to when I wiggle beta up by the same amount. I make beta a little bit bigger. Right. So I (Derivative of Tanx) make beta a little bit bigger by the same amount that I made alpha larger. And I think about how that affects the tangent of beta. The tangent of beta is this height divided by, divided by this width. And you can think about it, I mean this, this height maybe isn’t increasing a whole lot. But the width of this triangle is getting quite a bit narrower. And because it’s the ratio of that height to that width the tangent of the perturbed value of beta is quite a bit larger than the tangent of beta. And you can, you know it’s reflected in the fact the secant squared beta is a lot larger than secant squared alpha. So these, these kind of facts, right? The fact that the derivative of tangent is (Derivative of Tanx) secant squared theta you, you can really get a sense for why these things might be true by thinking about triangles and how wiggling the angle will affect certain ratios of sides of the triangles. But, if this seems a little too abstract we can kind of pull back a little bit and do do a numeric example next. You know, and maybe the numerical

example is sort of another

way to see a formula like this in action. Let’s do a numerical example to get a sense as to what you might do with the fact that the derivative of tangent is secen squared. Here’s an example, let’s try to approximate the value of tangent of 46 degrees. Why is this an interesting example? Well, we know the tangent of 45 degrees exactly, all right. And figure that out by looking at a triangle, here’s the angle, (Derivative of Tanx) 45 degrees, a right triangle, because the an, angles add up to, 180 degrees. So it’s 45 plus 90 plus what? Well, this must also be 45. It’s an isosceles triangle now, so these two sides are the same. A tangent is the ratio of this side to this side because they are equal that ratio was one. So I know the tangent of 45 exactly. It’s one.

But I am trying to figure out an approximation for the tangent of 46 degrees, the derivative tells me how wrigly an input affects the output, so I can use this fact and the fact that I know the derivative to try to approximate the tangent of 46. In particular the tangent of 46 degrees is the tangent of 45 plus one degree. And here you can see how I am perturbing the input a bit. And this is exactly (Derivative of Tanx) what the derivative would tell me something about. A little bit of bad news, the derivative of tangent is only secant squared if I do the measurement in radians. If I convert this to a problem in radians. With radians, says the tangent of pi/4, which is 45 degrees, plus with one degree in radians, is pi/180 radians, right? This is what I want to compute. I want to compute the tangent of pi/4 plus pi/180. And I’ll do that with approximation using the derivative. So according to the derivative this is about the tangent of pi over four, which is the tangent of 45 degrees, it’s one.

Plus the derivative of pi over four, which is secant squared pi over four. Times how much I wiggled the input by, which is pi over 180. I know the tangent of pi over four. It’s one. What’s the secant of pi over four squared? Well, if I pretend that these sides have length one, by the pythagorean theorem, this side must have length square root of two. The secant is hypotenuse over this width.

So the secant of pi over four is square root of two. So secant squared pi over four is square root of two squared, which is two times pi over 180. Now if you know an approximation for pi, you can compute two times pi over 180 plus one. And this is approximately 1.0349, and it keeps (Derivative of Tanx) going Pi’s irrational. And this is not so far off of the actual value. If we actually compute tangent of 46, the actual value is about 1.0355 and it keeps going. And· 1 0355. is awfully close to 1.0349. So we’ve successfully used the derivative to approximate the value of tangent 46 degrees.

# Derivative Proof of tan(x)

## Derivative proof of tan(x)

We can prove this derivative by using the derivatives of sin and cos, as well as quotient rule.

Write tangent in terms of sine and cosine

Take the derivative of both sides

Use Quotient Rule

Simplify

Use the Pythagorean identity for sine and cosine

and simplify

Derivative of Tanx

## Derivative proofs of csc(x), sec(x), and cot(x)

The derivative of these trig functions can be obtained easily from the Qoutient Rule using the reciprocals of sin(x), cos(x), and tan(x).

The basic trigonometric functions include the following 6 functions: sine (sinx), cosine (cosx), tangent (tanx), cotangent (cotx), secant (secx) and cosecant (cscx).

All these functions are continuous and differentiable in their domains. Below we make a list of derivatives for these functions.

### Derivatives of Basic Trigonometric Functions

We have already derived the derivatives of sine and cosine on the Definition of the Derivative page. They are as follows:

(sinx)=cosx,(cosx)=sinx.

Using the quotient rule it is easy to obtain an expression for the derivative of tangent:

(tanx)=(sinxcosx)=(sinx)cosxsinx(cosx)cos2x=cosxcosxsinx(sinx)cos2x=cos2x+sin2xcos2x=1cos2x.

The derivative of cotangent can be found in the same way. However, this can be also done using the chain rule for differentiating a composite function:

(cotx)=(1tanx)=1tan2x(tanx)=1sin2xcos2x1cos2x=cos2xsin2xcos2x=1sin2x.

Similarly, we find the derivatives of secant and cosecant:

(secx)=(1cosx)=1cos2x(cosx)=sinxcos2x=sinxcosx1cosx=tanxsecx,

(cscx)=(1sinx)=1sin2x(sinx)=cosxsin2x=cosxsinx1sinx=cotxcscx.

Derivative of Tanx

## Derivative of Tanx Solved Problems

Click or tap a problem to see the solution.

y=cos2x2sinx

y=tanx+13tan3x

y=cosx13cos3x

y=1cosnx

y=sinx1+cosx

y=cos2sinx

y=xsinx+cosx

y=sin2x

y=cos1x

y=sin3x+cos3x

y=tanx2cotx2

### Example 12

y=x2sinx+2xcosx2sinx

y=tan2x+lncos2x

y=sinnxcosnx

y=ln1sinx1+sinx

### Example 16

Calculate the derivative of the functiony=(2x2)cosx+2xsinxat x=π.

### Example 17

Calculate the derivative of the functiony=(x+1)cosx+(x+2)sinxat x=0.

y=sec2x2+csc2x2

### Example 19

y=(tanx)cosx,where0<x<π2.

### Example 20

y=sin2x1+cotx+cos2x1+tanx