微分積分学‎:
微分: 積の計算

1.  
ddxexcos⁡x\frac{d}{dx} {e}^{x}\cos{x}
  解答
2.  
ddxsin⁡xcos⁡x\frac{d}{dx} \sin{x}\cos{x}
  解答
3.  
ddxxcos⁡x\frac{d}{dx} x\cos{x}
  解答
4.  
ddxxexcos⁡x\frac{d}{dx} x{e}^{x}\cos{x}
  解答
5.  
ddxln⁡xx\frac{d}{dx} \ln{x}x
  解答
6.  
ddx7xtan⁡x\frac{d}{dx} 7x\tan{x}
  解答
7.  
ddx1exx3\frac{d}{dx} \frac{1}{{e}^{x}{x}^{3}}
  解答
8.  
ddxx2log⁡x\frac{d}{dx} {x}^{2}\log{x}
  解答

Product Rule for Derivatives - Introduction

In calculus, students are often tasked with finding the “derivative” of a given function. The derivative of a function
y=f(x)y=f(x) of a variable
xx
is a measure of the rate at which the value of the function, which is
yy
, changes with respect to the change of the variable
xx
. We call this the derivative of
ff with respect to
xx
.
What happens if we’re asked to find the derivative of the expression
excos⁡x{e}^{x}\cos{x}
?
This might seem like a daunting task, since this expression contains both an exponential function and a trigonometric function. Thankfully, we have a set of derivative rules that can help us find the derivative of most functions easily. Since the expression
excos⁡x{e}^{x}\cos{x}
is a product of
ex{e}^{x}
and
cos⁡x\cos{x}
, we can use the product rule in this case, which states:
(fg)′=f′g+fg′(fg)'=f'g+fg'
Let’s see how this works in the example problem above.

Using Product Rule for Derivatives

In case you are not familiar with all the notations, there are two main ways to indicate the derivative of a function:
1)
ddx\frac{d}{dx}
where
xx
is the "with respect to" variable
2) Just an apostrophe, like
f′(x)f'(x), or simply
f′f'
In the example
ddxexcos⁡x\frac{d}{dx} {e}^{x}\cos{x} above, the
ddx\frac{d}{dx}
notation tells us that we’re looking for the derivative of
excos⁡x{e}^{x}\cos{x}
. The product rule, meanwhile, says that the derivative of
fgfg
is equal to the derivative of
ff multiplied by
gg
, plus
ff
multiplied by the derivative of
gg
.
In order to use the derivative product rule, as with any rule in calculus, first we need to assign parts of our expression to the appropriate variables in the rule. In this case, they are
ff
and
gg
. Let
f=exf={e}^{x}
and
g=cos⁡xg=\cos{x}
.
Now, we can use the product rule to give:
ddxexcos⁡x=(ddxex)cos⁡x+ex(ddxcos⁡x)\frac{d}{dx} {e}^{x}\cos{x}=(\frac{d}{dx} {e}^{x})\cos{x}+{e}^{x}(\frac{d}{dx} \cos{x})
Using other derivative rules that we will not go into detail here, we know that the derivative of
ex{e}^{x}
is
ex{e}^{x}
, and the derivative of
cos⁡x\cos{x}
is
−sin⁡x-\sin{x}
.
We are done. We have found the derivative:
ddxexcos⁡x=excos⁡x−exsin⁡x\frac{d}{dx} {e}^{x}\cos{x}={e}^{x}\cos{x}-{e}^{x}\sin{x}

Getting More Practice

The best way to become familiar with these rules? Try out more practice problems at the top of this page! Once you are comfortable with the product rule, you can also try other practice problems. Over time, hopefully you will find derivative problems no longer overwhelming.
At Cymath, we believe that regular math practice combined with step-by-step explanations can help students gain confidence and command of most differentiation and integration problems. Get started today with our online help, or download the Cymath homework helper app for iOS and Android today!
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