Product Rule - Knight Math
2C: Product and Quotient RuleSo, we have specific rules for the differentiation of power functions and some trig. functions. We also have general rules for the derivatives of constant multiples of functions and the sum or difference of some functions. Now we need to investigate the case where we have a product or quotient of two functions.Product RuleFirst, we need to figure out how to find the derivative of functions like fx=x2sinx.Let’s make a Rule!Ok, let’s write our derivative as u'=limΔx→0Δu-uΔx, and v'=limΔx→0Δv-vΔxNow let’s look at the derivative ddxuv for functions ux and vx.uv'=limΔx→0ΔuΔv-uvΔx-469901841500Product RuleIf u and v are differentiable functions, then uv is a differentiable function and ddxuv=u'v+uv'That is, the derivative of a product is equal to "The derivative of the 1st times the 2nd, plus the 1st times the derivative of the 2nd." 00Product RuleIf u and v are differentiable functions, then uv is a differentiable function and ddxuv=u'v+uv'That is, the derivative of a product is equal to "The derivative of the 1st times the 2nd, plus the 1st times the derivative of the 2nd." Now, I wish this fraction said Δu-uΔx or Δv-vΔx so we could write it as u' and v'. Let’s get creative…uv'=limΔx→0ΔuΔv-uvΔx=limΔx→0ΔuΔv-uΔv+uΔv-uvΔx=limΔx→0Δu-uΔv+uΔv-vΔxlimΔx→0Δu-uΔxΔv+u(Δv-v)Δx=u'v+uv' Examples Find the derivative of the function. fx=x3(5x5-x2)gx=5x3cosxhx=2xcosx-2sinxQuotient RuleIf multiplication gets its own rule, division should get one too. Let’s find a rule for division quotients.-25401424305Quotient RuleIf u and v are differential functions, so is uv', and uv'=u'v-uv'v2That is, the derivative of a quotient is equal to "The derivative of the top times the bottom,minus the top times the derivative of the bottom,all over the bottom squared." 00Quotient RuleIf u and v are differential functions, so is uv', and uv'=u'v-uv'v2That is, the derivative of a quotient is equal to "The derivative of the top times the bottom,minus the top times the derivative of the bottom,all over the bottom squared." So, we start with two differentiable functions u(x) and v(x).uv'=limΔx→01ΔxΔuΔv-uv=limΔx→01ΔxΔuv-uΔvΔv?v=limΔx→01ΔxΔuv-uv-uΔv+uvΔv?v=limΔx→01ΔxΔu-uv-uΔv-vΔv?v=limΔx→0Δu-uΔxv-uΔv-vΔxΔv?v=u'v-uv'v2Examples Use the quotient rule to find these derivatives.ddx4x2-1x+3ddxsinxcosxPutting it all together… Combo time!Now we have two very useful tools, let’s put them together to find the derivative of some more complex functions.Try it! Find the derivative of the following functions.y=xsinxx+1gx=xx+1(sinx)How about some more trig. functions…ddθtanθ=ddθsinθcosθ ddθcotθ=ddθcosθsinθ ddθcscθ=ddθ1sinθ ddθsecθ=ddθ1cosθ 46672523431500Here’s how the AP test will use these rules:Higher DerivativesNow that we have more ways to find derivatives, we can start to find higher derivatives. This just means we will take the derivative of the derivative.Example Find the second derivative of st=-.81t2+2 which is the position function for a falling object on the moon.228219029083000When working with physics applications, we have a very important relationship that we will work more with later:165544527876500In General, this table gives us the notation for higher derivatives: ................
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