微分積分学‎:
微分: 三角関数

1.  
ddxx4tan⁡x\frac{d}{dx} {x}^{4}\tan{x}
  解答
2.  
ddx4cos⁡x−3sin⁡x\frac{d}{dx} 4\cos{x}-3\sin{x}
  解答
3.  
ddxcsc⁡xcot⁡x\frac{d}{dx} \csc{x}\cot{x}
  解答
4.  
ddxsin⁡2xcos⁡2x\frac{d}{dx} \frac{\sin^{2}x}{\cos^{2}x}
  解答
5.  
ddxcot⁡5x\frac{d}{dx} \cot^{5}x
  解答
6.  
ddxcsc⁡xx\frac{d}{dx} \csc{x}x
  解答
7.  
ddx3tan⁡x+sec⁡x\frac{d}{dx} 3\tan{x}+\sec{x}
  解答
8.  
ddx1tan⁡x\frac{d}{dx} \frac{1}{\tan{x}}
  解答
9.  
ddxsin⁡2xcos⁡3x\frac{d}{dx} \sin^{2}x\cos^{3}x
  解答

Derivatives of Trigonometric Functions - Introduction

By now, you should have seen the derivatives of basic functions such as polynomials. We will now start exploring the derivatives of trigonometric functions. First, let us list the rules:
ddxsin⁡x=cos⁡x\frac{d}{dx} \sin{x}=\cos{x}

ddxcos⁡x=−sin⁡x\frac{d}{dx} \cos{x}=-\sin{x}

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

ddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx} \csc{x}=-\csc{x}\cot{x}

ddxsec⁡x=sec⁡xtan⁡x\frac{d}{dx} \sec{x}=\sec{x}\tan{x}

ddxcot⁡x=−csc⁡2x\frac{d}{dx} \cot{x}=-\csc^{2}x
Are you curious about how these rules were derived? Let's explore this in the next section. Note that these rules are also on our reference page on trigonometric differentiation.

Derivatives of Trigonometric Functions - Proof

Let's see if we can prove that
ddxtan⁡x=sec⁡2x\frac{d}{dx} \tan{x}=\sec^{2}x
.
First, we start with the left hand side of the equation:
ddxtan⁡x\frac{d}{dx} \tan{x}
By the trigonometric identity of
tan⁡x=sin⁡xcos⁡x\tan{x}=\frac{\sin{x}}{\cos{x}}
, we have:
ddxtan⁡x=ddxsin⁡xcos⁡x\frac{d}{dx} \tan{x}=\frac{d}{dx} \frac{\sin{x}}{\cos{x}}
Then, we apply the quotient rule:
ddxtan⁡x=cos⁡xcos⁡x−sin⁡x(−sin⁡x)cos⁡2x\frac{d}{dx} \tan{x}=\frac{\cos{x}\cos{x}-\sin{x}(-\sin{x})}{\cos^{2}x}
Simplify:
ddxtan⁡x=cos⁡2x+sin⁡2xcos⁡2x\frac{d}{dx} \tan{x}=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}
Apply the trigonometric identity of
cos⁡2x+sin⁡2x=1\cos^{2}x+\sin^{2}x=1
:
ddxtan⁡x=1cos⁡2x\frac{d}{dx} \tan{x}=\frac{1}{\cos^{2}x}
Apply the trigonometric identity of
1cos⁡x=sec⁡x\frac{1}{\cos{x}}=\sec{x}
:
ddxtan⁡x=sec⁡2x\frac{d}{dx} \tan{x}=\sec^{2}x
We are done. We have shown that the left hand side equals the right hand side, and that the derivative of
tan⁡x\tan{x}
is indeed
sec⁡2x\sec^{2}x
.

What's Next

A good way to get better at finding derivatives for trigonometric functions is more practice! You can try out more practice problems at the top of this page. Once you are familiar with this topic, you can also try other practice problems. Soon, you will find all derivatives problems easy to solve.
At Cymath, we believe that sufficient practice and step-by-step guidance can help students master most differentiation and integration problems. You can try our online solver anytime, or download the Cymath homework helper app for iOS and Android today!
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