TutorTube: Derivative Rules Spring 2020 Introduction Hello and welcome to TutorTube, where The Learning Center’s Lead Tutors help you understand challenging course concepts with easy to understand videos. My name is Haley Higginbotham, Lead Tutor for Math. In today’s video, we will explore derivative rules. Let’s get started! Notation Notes

in front of a function means "take the derivative." 𝑑𝑑 𝑑𝑑𝑑𝑑( ) = means that whatever is after the equals sign, is the derivative of ( ). ′ 𝑓𝑓 𝑥𝑥 𝑓𝑓 𝑥𝑥 Constant Rule

= 𝒅𝒅 𝒄𝒄 𝟎𝟎 The constant rule says that when you𝒅𝒅𝒅𝒅 take the derivative of a constant, it is . We will see why this is in a second, while we talk about power rule. 𝟎𝟎

Power Rule

= 𝒅𝒅 𝒏𝒏 𝒏𝒏−𝟏𝟏 𝒙𝒙 𝒏𝒏𝒙𝒙 The power rule says that if you have𝒅𝒅𝒅𝒅 to some power , the derivative is the power times raised to the power minus . 𝒙𝒙 𝒏𝒏 𝒙𝒙 𝟏𝟏 Examples:

1. 𝒅𝒅 You𝒅𝒅𝒅𝒅 might𝟓𝟓 be confused on how to use the power rule on since it doesn't have an raised to a power. However, it actually does. has an next to it, since 𝟓𝟓 anything raised to the zero power is , therefore = . So, using𝟎𝟎 the power rule 𝒙𝒙 𝟓𝟓 𝒙𝒙 gives us ( ) = = . This is why the constant rule𝟎𝟎 is true for any constant. 𝟎𝟎−𝟏𝟏 −𝟏𝟏 𝟏𝟏 𝟓𝟓 𝟓𝟓𝒙𝒙 𝟓𝟓 𝟎𝟎𝒙𝒙 𝟎𝟎𝒙𝒙 𝟎𝟎

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2. 𝒅𝒅 𝟑𝟑 𝒅𝒅𝒅𝒅 𝟕𝟕𝒙𝒙 Using the power rule, we have ( ) = . 𝟑𝟑−𝟏𝟏 𝟐𝟐 So, = 𝟕𝟕. 𝟑𝟑𝒙𝒙 𝟐𝟐𝟏𝟏𝒙𝒙 𝒅𝒅 𝟑𝟑 𝟐𝟐 𝒅𝒅𝒅𝒅 𝟕𝟕𝒙𝒙 𝟐𝟐𝟐𝟐𝒙𝒙 3. Find the derivative of ( ) = + . 𝟑𝟑 We can use the power rule𝒇𝒇 𝒙𝒙 on 𝟔𝟔each𝒙𝒙 − 𝟗𝟗individual𝟗𝟗 𝟒𝟒 term since we are adding and subtracting. Using the power rule on each term, we get ( ) ( ) + ( ) 𝟑𝟑−𝟏𝟏 𝟏𝟏−𝟏𝟏 𝟎𝟎−𝟏𝟏 𝟔𝟔 𝟑𝟑𝒙𝒙 = − 𝟗𝟗 𝟏𝟏𝒙𝒙 + 𝟒𝟒 𝟎𝟎 𝒙𝒙 𝟐𝟐 𝟎𝟎 −𝟏𝟏 𝟏𝟏𝟏𝟏=𝒙𝒙 − 𝟗𝟗𝒙𝒙 𝟎𝟎𝒙𝒙 𝟐𝟐 Practice: 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟗𝟗 Find the derivative. 1. 5 4 2. 2𝑥𝑥 + 4 7 + 140 7 4 3. 𝑥𝑥 + 5 𝑥𝑥 − 𝑥𝑥

√𝑥𝑥 𝑥𝑥 Log and Exponential Function Rules General Logarithm Rule:

( ) = ( ) ( ) ( ) 𝒅𝒅 𝟏𝟏 ′ �𝒍𝒍𝒍𝒍𝒍𝒍𝒂𝒂�𝒇𝒇 𝒙𝒙 �� ⋅ 𝒇𝒇 𝒙𝒙 Special case: 𝒅𝒅𝒅𝒅 𝒇𝒇 𝒙𝒙 𝒍𝒍𝒍𝒍 𝒂𝒂 ( ) ( ) = ′( ) 𝒅𝒅 𝒇𝒇 𝒙𝒙 �𝒍𝒍𝒍𝒍�𝒇𝒇 𝒙𝒙 �� 𝒅𝒅𝒅𝒅 𝒇𝒇 𝒙𝒙 Super special case:

[ ( )] = 𝒅𝒅 𝟏𝟏 𝒍𝒍𝒍𝒍 𝒙𝒙 𝒅𝒅𝒅𝒅 𝒙𝒙 3

General Exponential Rule:

( ) = ( ) ( ) ( ) 𝒅𝒅 𝒇𝒇 𝒙𝒙 𝒇𝒇 𝒙𝒙 ′ �𝒂𝒂 � 𝒂𝒂 ⋅ 𝒍𝒍𝒍𝒍 𝒂𝒂 ⋅ 𝒇𝒇 𝒙𝒙 Special case: 𝒅𝒅𝒅𝒅

( ) = ( ) ( ) 𝒅𝒅 𝒇𝒇 𝒙𝒙 𝒇𝒇 𝒙𝒙 ′ �𝒆𝒆 � 𝒆𝒆 ⋅ 𝒇𝒇 𝒙𝒙 𝒅𝒅𝒅𝒅 Super special case:

[ ] = 𝒅𝒅 𝒙𝒙 𝒙𝒙 𝒆𝒆 𝒆𝒆 𝒅𝒅𝒅𝒅 These are the typical exponential and logarithm derivatives used in calculus. The general exponential and logarithm derivatives come from using the limit definition of the derivative to find the derivative. The special cases are actually the general rules in disguise. If we apply the general log rule to natural log of ( ), we get ( ) 𝒇𝒇 𝒙𝒙 ( ) = ( ) = ( ) ( ) ′( ) 𝒅𝒅 𝟏𝟏 ′ 𝒇𝒇 𝒙𝒙 �𝒍𝒍𝒍𝒍�𝒇𝒇 𝒙𝒙 �� ⋅ 𝒇𝒇 𝒙𝒙 which is our special case𝒅𝒅𝒅𝒅 for logs. Then,𝒇𝒇 𝒙𝒙 if⋅ we𝒍𝒍𝒍𝒍 𝒆𝒆apply the general𝒇𝒇 𝒙𝒙 rule for log to natural log of , we get

𝒙𝒙 [ ( )] = = ( ) 𝒅𝒅 𝟏𝟏 𝟏𝟏 𝒍𝒍𝒍𝒍 𝒙𝒙 ⋅ 𝟏𝟏 𝒅𝒅𝒅𝒅 𝒙𝒙 ⋅ 𝒍𝒍𝒍𝒍 𝒆𝒆 𝒙𝒙 which is our super special case for logs. Similarly, if the general rule for exponentials is applied to raised to the ( ) we get

(𝒆𝒆) = ( ) ( )𝒇𝒇 𝒙𝒙 ( ) = ( ) ( ) 𝒅𝒅 𝒇𝒇 𝒙𝒙 𝒇𝒇 𝒙𝒙 ′ 𝒇𝒇 𝒙𝒙 ′ �𝒆𝒆 � 𝒆𝒆 ⋅ 𝒍𝒍𝒍𝒍 𝒆𝒆 ⋅ 𝒇𝒇 𝒙𝒙 𝒆𝒆 ⋅ 𝒇𝒇 𝒙𝒙 which is our special case𝒅𝒅𝒙𝒙 for exponentials. If we apply the general rule for exponentials to raised to the we get

𝒆𝒆 [𝒙𝒙 ] = ( ) = 𝒅𝒅 𝒙𝒙 𝒙𝒙 𝒙𝒙 𝒆𝒆 𝒆𝒆 ⋅ 𝒍𝒍𝒍𝒍 𝒆𝒆 ⋅ 𝟏𝟏 𝒆𝒆 which is our super special case𝒅𝒅𝒅𝒅 for exponentials. 4

Product Rule

[ ( ) ( )] = ( ) ( ) + ( ) ( ) 𝒅𝒅 ′ ′ 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝒅𝒅𝒙𝒙 Examples: 1. Find the derivative of ( ) = ( + )( + ). 𝟐𝟐 𝟑𝟑 𝟐𝟐 Now, if we just had the power𝒉𝒉 𝒙𝒙 rule,𝟔𝟔𝒙𝒙 we− 𝟕𝟕 would𝟕𝟕 𝟗𝟗 have𝒙𝒙 𝟒𝟒 to𝒙𝒙 multiply− 𝟏𝟏𝟏𝟏𝟏𝟏 − everything𝟑𝟑 out and then take the derivative of each individual term, which seems like a lot of work. Luckily, we have the product rule to make things a little easier on us. So let's go ahead and label what is going to be ( ) and what is going to be ( ).

To match our product rule, let's have 𝒇𝒇+𝒙𝒙 = ( ) and + 𝒈𝒈 𝒙𝒙 = ( ). The product rule says that we also𝟐𝟐 need the derivatives𝟑𝟑 of (𝟐𝟐 ) and ( ). So let's find those derivatives using𝟔𝟔𝒙𝒙 the− 𝟕𝟕 power𝟕𝟕 𝟗𝟗 rule.𝒇𝒇 𝒙𝒙 𝒙𝒙 𝟒𝟒𝒙𝒙 − 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟑𝟑 𝒈𝒈 𝒙𝒙 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 ( ) = + ( ) = ( ) = + 𝟐𝟐 ( )′ = + 𝒇𝒇 𝒙𝒙 𝟑𝟑 𝟔𝟔𝒙𝒙 −𝟐𝟐 𝟕𝟕𝟕𝟕 𝟗𝟗 ′ 𝒇𝒇 𝒙𝒙 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟕𝟕 𝒈𝒈 𝒙𝒙 𝒙𝒙 𝟒𝟒𝒙𝒙 − 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟑𝟑 𝒈𝒈 𝒙𝒙 𝟑𝟑𝒙𝒙 𝟖𝟖𝟖𝟖 − 𝟏𝟏𝟏𝟏 Now that we have all the pieces, let's put them together. ( ) ( ) + ( ) ( ) ′ ′ ( ) = ( )( + 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝒇𝒇) +𝒙𝒙 (𝒈𝒈 𝒙𝒙 + )( + ) ′ 𝟑𝟑 𝟐𝟐 𝟐𝟐 𝟐𝟐 You could𝒉𝒉 𝒙𝒙 multiply𝟏𝟏𝟏𝟏𝟏𝟏 everything− 𝟕𝟕 𝒙𝒙 𝟒𝟒 𝒙𝒙out,− but𝟏𝟏𝟏𝟏𝟏𝟏 usually− 𝟑𝟑 it𝟔𝟔 𝒙𝒙is acceptable− 𝟕𝟕𝟕𝟕 𝟗𝟗 𝟑𝟑𝒙𝒙 to just𝟖𝟖𝒙𝒙 leave− 𝟏𝟏𝟏𝟏 it as shown above. However, it will depend on what your specific professor wants.

2. Find the derivative of ( ) = ( + ) 𝟐𝟐 𝟐𝟐𝒙𝒙 𝟑𝟑 For this example, even multiplying𝒋𝒋 𝒙𝒙 𝒆𝒆 out𝟒𝟒𝒙𝒙 wouldn't𝟕𝟕 help since you would still end up having to do the product rule, except you would have to do it twice and not just once. Like before, let's label our different functions.

( ) = ( ) = 𝟐𝟐 ( ) = 𝟐𝟐𝒙𝒙+ ′ ( ) = 𝒇𝒇 𝒙𝒙 𝒇𝒇 𝒙𝒙 𝒆𝒆𝟑𝟑 ′ 𝒈𝒈 𝒙𝒙 𝟒𝟒𝒙𝒙 𝟕𝟕 𝒈𝒈 𝒙𝒙 Now, let's find the derivatives.

( ) = ( ) = 𝟐𝟐 𝟐𝟐 ( ) = 𝟐𝟐𝒙𝒙+ ′ ( ) = 𝟐𝟐𝒙𝒙 𝒇𝒇 𝒙𝒙 𝒆𝒆𝟑𝟑 𝒇𝒇 ′𝒙𝒙 𝟒𝟒𝟒𝟒𝒆𝒆 𝟐𝟐 𝒈𝒈 𝒙𝒙 𝟒𝟒𝒙𝒙 𝟕𝟕 𝒈𝒈 𝒙𝒙 𝟏𝟏𝟏𝟏𝒙𝒙 5

Putting it all together:

( ) ( ) + ( ) ( ) ′ ′ ( ) = 𝒇𝒇 𝒙𝒙 𝒈𝒈(𝒙𝒙 +𝒇𝒇 𝒙𝒙) +𝒈𝒈 𝒙𝒙 ( ) 𝟐𝟐 𝟐𝟐 ′ 𝟐𝟐𝒙𝒙 𝟑𝟑 𝟐𝟐𝒙𝒙 𝟐𝟐 Practice: 𝒋𝒋 𝒙𝒙 �𝟒𝟒𝟒𝟒𝒆𝒆 � 𝟒𝟒𝒙𝒙 𝟕𝟕 �𝒆𝒆 � 𝟏𝟏𝟏𝟏𝒙𝒙 Find the derivative. 4. ( + 3)( 2 + 5) 2 5. 𝑥𝑥 ( +𝑥𝑥6)− 𝑥𝑥 3 6. √𝑥𝑥( 𝑥𝑥)(5 3 ) 3 2 𝑙𝑙𝑙𝑙 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 Quotient Rule ( ) ( ) ( ) ( ) ( ) = ( ) ′ [ ( )] ′ 𝒅𝒅 𝒇𝒇 𝒙𝒙 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 − 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 � � 𝟐𝟐 𝒅𝒅𝒅𝒅 𝒈𝒈 𝒙𝒙 𝒈𝒈 𝒙𝒙 Memorization Tip: The top of the quotient rule is the same as the product rule, just with a minus sign instead of a plus sign.

Examples:

( ) 1. Find the derivative of ( ) = 𝒍𝒍𝒍𝒍 𝒙𝒙 Similar to the product rule,𝒉𝒉 𝒙𝒙we are𝟐𝟐𝟐𝟐 going to label, find the derivatives of those labelled pieces, then plug it into our rule. ( ) = ( ) ( ) = 𝒇𝒇 𝒙𝒙( ) =𝒍𝒍𝒍𝒍 𝒙𝒙 ′ ( ) = 𝟏𝟏 𝒇𝒇 𝒙𝒙 ′ 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝟐𝟐𝟐𝟐 𝒈𝒈 𝒙𝒙 𝟐𝟐 ( ) ( ) ( ) ( )

′ [ ( )] ′ 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 − 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝟐𝟐 𝒈𝒈 𝒙𝒙 6

( ) ( ( ))( ) ( ) = 𝟏𝟏 ( ) ′ �𝒙𝒙� 𝟐𝟐𝟐𝟐 − 𝒍𝒍𝒍𝒍 𝒙𝒙 𝟐𝟐 𝒉𝒉 𝒙𝒙 𝟐𝟐( ) ( ) = 𝟐𝟐𝟐𝟐 ′ 𝟐𝟐 − 𝟐𝟐 𝒍𝒍𝒍𝒍 𝒙𝒙 𝒉𝒉 𝒙𝒙 𝟐𝟐 Typically, you only have to simplify your answer𝟒𝟒𝒙𝒙 if the question says to, or if it's not too much trouble to simplify it.

( ) = 2. Find the derivative of 𝟑𝟑 𝟒𝟒𝒙𝒙 −𝟗𝟗𝟗𝟗 𝟐𝟐 ( ) = 𝒋𝒋 𝒙𝒙 𝟓𝟓𝒙𝒙 +𝟐𝟐 ( ) = ( ) = 𝟑𝟑 + ′ ( ) = 𝟐𝟐 𝒇𝒇 𝒙𝒙 𝟒𝟒𝒙𝒙 𝟐𝟐− 𝟗𝟗𝟗𝟗 𝒇𝒇 𝒙𝒙′ 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟗𝟗 𝒈𝒈 𝒙𝒙 𝟓𝟓𝒙𝒙 𝟐𝟐 𝒈𝒈 𝒙𝒙 𝟏𝟏𝟏𝟏𝟏𝟏 ( ) ( ) ( ) ( )

′ [ ( )] ′ 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 − 𝒇𝒇 𝒙𝒙 𝒈𝒈 𝒙𝒙 𝟐𝟐 ( )( 𝒈𝒈 𝒙𝒙+ ) ( )( ) ( ) = 𝟐𝟐 𝟐𝟐( + ) 𝟑𝟑 ′ 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟗𝟗 𝟓𝟓𝒙𝒙 𝟐𝟐 − 𝟒𝟒𝒙𝒙 − 𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏𝟏𝟏 𝒋𝒋 𝒙𝒙 𝟐𝟐 𝟐𝟐 ( )( 𝟓𝟓𝒙𝒙+ )𝟐𝟐 + ( ) = 𝟐𝟐 ( 𝟐𝟐 + ) 𝟒𝟒 𝟐𝟐 ′ 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟗𝟗 𝟓𝟓𝒙𝒙 𝟐𝟐 − 𝟒𝟒𝟒𝟒𝒙𝒙 𝟗𝟗𝟗𝟗𝒙𝒙 𝒋𝒋 𝒙𝒙 𝟐𝟐 𝟐𝟐 Practice: 𝟓𝟓𝒙𝒙 𝟐𝟐 Find the derivative.

7. 3𝑥𝑥+9 2−𝑥𝑥 8. 2 𝑥𝑥 +4𝑥𝑥+20 9. 3− 𝑥𝑥 𝑥𝑥+5 3 𝑥𝑥 −7 Chain Rule

( ) = ( ) ( ) 𝒅𝒅 ′ ′ �𝒇𝒇�𝒈𝒈 𝒙𝒙 �� 𝒇𝒇 �𝒈𝒈 𝒙𝒙 � ⋅ 𝒈𝒈 𝒙𝒙 Examples: 𝒅𝒅𝒅𝒅 1. (3 6 ) 𝑑𝑑 3 2 8 𝑑𝑑𝑑𝑑 𝑥𝑥 − 𝑥𝑥 7

Similar to the product and quotient rules, we are going to label and take the individual derivatives then put it into the chain rule. However, it can be a little confusing at first to figure out what should be labelled as what, but after some practice it becomes clearer. Typically, when talking about the chain rule, there are two parts. The "inside" ( ( )) and the "outside" ( ( )). The "inside" usually refers to whatever is being raised to a power or inside the square root. The 𝒈𝒈 𝒙𝒙 𝒇𝒇 𝒙𝒙 "outside" is usually the power or the square root. In this example, the "inside" is since that is the function being raised to a power. The "outside" is the power𝟑𝟑 of 𝟐𝟐 . To take the derivative, we leave what's 𝟑𝟑𝒙𝒙 − 𝟔𝟔𝒙𝒙 on the inside the same and take the derivative of the outside function, then 𝟖𝟖 multiply by the derivative of the inside.

= [ ] ( ) 𝟑𝟑 𝟐𝟐 𝟕𝟕 𝒅𝒅 𝟑𝟑 𝟐𝟐 𝟖𝟖 𝟑𝟑𝒙𝒙( − 𝟔𝟔𝒙𝒙 ⋅) ( 𝟑𝟑𝒙𝒙 − 𝟔𝟔)𝒙𝒙 = 𝒅𝒅𝒅𝒅 𝟑𝟑 𝟐𝟐 𝟕𝟕 = 𝟖𝟖( 𝟑𝟑𝒙𝒙 − 𝟔𝟔𝒙𝒙 )( 𝟗𝟗𝟗𝟗 − 𝟏𝟏𝟏𝟏𝟏𝟏) 𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟕𝟕 =𝟖𝟖( 𝟗𝟗𝒙𝒙 − 𝟏𝟏𝟏𝟏𝟏𝟏)( 𝟑𝟑𝒙𝒙 − 𝟔𝟔𝒙𝒙 ) 𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟕𝟕 2. Find the derivative of = 𝟕𝟕𝟕𝟕+𝒙𝒙 − 𝟗𝟗𝟗𝟗 𝟑𝟑𝒙𝒙 − 𝟔𝟔𝒙𝒙 𝟐𝟐 𝒚𝒚 √𝟑𝟑𝒙𝒙 𝟓𝟓 ( + ) Rewrite so that it looks like something we are more familiar with: 𝟏𝟏. Now it looks similar to the last example. 𝟐𝟐 𝟐𝟐 𝟑𝟑𝒙𝒙 𝟓𝟓 = [ + ] ( + ) 𝟏𝟏 −𝟏𝟏 ′ 𝟏𝟏 𝟐𝟐 𝟐𝟐 𝒅𝒅 𝟐𝟐 𝒚𝒚 𝟑𝟑𝒙𝒙 𝟓𝟓 ⋅ 𝟑𝟑𝒙𝒙 𝟓𝟓 𝟐𝟐 = ( + )𝒅𝒅𝒙𝒙( ) 𝟏𝟏 − ′ 𝟏𝟏 𝟐𝟐 𝟐𝟐 𝒚𝒚 𝟑𝟑𝒙𝒙 𝟓𝟓 𝟔𝟔𝟔𝟔 𝟐𝟐 = + ′ 𝟑𝟑𝟑𝟑 𝒚𝒚 𝟐𝟐 Practice: √𝟑𝟑𝒙𝒙 𝟓𝟓 Find the derivative. 10. (2 + 5) 3 11. 𝑥𝑥 4 + 8 3 12. √𝑥𝑥 − 𝑥𝑥 3𝑥𝑥−2 𝑒𝑒

8

Answers to Practice Problems: 1. 20 3 2. 14𝑥𝑥 + 16 7 6 3 3. 𝑥𝑥+ 5 𝑥𝑥 − 1 4. 2(√𝑥𝑥 2 + 5) + ( + 3)(2 2) which when simplified fully, gives us 3 + 1 2 2 5. 𝑥𝑥 (− +𝑥𝑥 6) + 3 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 𝑥𝑥 − 1 3 2 6. 2(√5𝑥𝑥 𝑥𝑥 3 ) + (𝑥𝑥 )√(𝑥𝑥15 6 ) 2 2 7. ( 𝑥𝑥 ) − 𝑥𝑥 𝑙𝑙𝑙𝑙 𝑥𝑥 𝑥𝑥 − 𝑥𝑥 −3 2 2−𝑥𝑥 8. (2 ) −𝑥𝑥 +6𝑥𝑥+32 2 3−𝑥𝑥 9. (3 ) 2 −2𝑥𝑥 −15𝑥𝑥 −7 3 2 10. 6(2𝑥𝑥 −+7 5) 2 𝑥𝑥 11. 2 3𝑥𝑥 −4 3 12. 23√𝑥𝑥 −4𝑥𝑥 +8 3𝑥𝑥−2 Outro𝑒𝑒 Thank you for watching TutorTube! I hope you enjoyed this video. Please subscribe to our channel for more exciting videos. Check out the links in the description below for more information about The Learning Center and follow us on social media. See you next time!