Math 1151, Calculus ILecture XXIIIReview for Midterm IIMath 1151, Calculus I Lecture XXIIIReview for Midterm II 1 / 9Midterm remarksThe midterm will cover stuff through Section 4.1——————————————————————————-There has been some confusion regarding exam times, because the startingtime listed on the Math 1151 website is now “7:40pm”, but on the coursecalendar it says “7:30pm”.Students should arrive at the exam location at 7:30pm. Exams will bedistributed between 7:30 and 7:40pm. the exam will start at 7:40pmsharp, and go to 8:35pm.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 2 / 9Zombies and related ratesImaging two players in this zombie outbreak game I see advertised oncampus.A zombie is running in the positive direction in the x axis staring from(0, 0) at a rate of 1 meter/second.The other player standing at the point (0, −3) fires a Nerf dart at him.The dart travels at a rate of 4 meters/second in the positive y-direction.After 1 second how fast is the dart approaching / traveling away from thezombieMath 1151, Calculus I Lecture XXIIIReview for Midterm II 3 / 9Compute some derivativesCompute f0(x) where f (x) = cos(x2− x)f (x) = u(v(x)) where v (x) = x2− x and u(x) = cos(x).By the chain rule: f0(x) = u0(v(x))v0(x)u0(x) = −sin(x), v0(x) = 2x − 1Substitute these inf0(x) = − sin(x2− x) · (2x − 1)Math 1151, Calculus I Lecture XXIIIReview for Midterm II 4 / 9Compute some derivativesCompute f0(x) where f (x) = cos(x2− x)f (x) = u(v(x)) where v (x) = x2− x and u(x) = cos(x).By the chain rule: f0(x) = u0(v(x))v0(x)u0(x) = −sin(x), v0(x) = 2x − 1Substitute these inf0(x) = − sin(x2− x) · (2x − 1)Math 1151, Calculus I Lecture XXIIIReview for Midterm II 4 / 9Compute some derivativesCompute f0(x) where f (x) = cos(x2− x)f (x) = u(v(x)) where v (x) = x2− x and u(x) = cos(x).By the chain rule: f0(x) = u0(v(x))v0(x)u0(x) = −sin(x), v0(x) = 2x − 1Substitute these inf0(x) = − sin(x2− x) · (2x − 1)Math 1151, Calculus I Lecture XXIIIReview for Midterm II 4 / 9Compute some derivativesCompute f0(x) where f (x) = cos(x2− x)f (x) = u(v(x)) where v (x) = x2− x and u(x) = cos(x).By the chain rule: f0(x) = u0(v(x))v0(x)u0(x) = −sin(x), v0(x) = 2x − 1Substitute these inf0(x) = − sin(x2− x) · (2x − 1)Math 1151, Calculus I Lecture XXIIIReview for Midterm II 4 / 9Compute some derivativesCompute f0(x) where f (x) = cos(x2− x)f (x) = u(v(x)) where v (x) = x2− x and u(x) = cos(x).By the chain rule: f0(x) = u0(v(x))v0(x)u0(x) = −sin(x), v0(x) = 2x − 1Substitute these inf0(x) = − sin(x2− x) · (2x − 1)Math 1151, Calculus I Lecture XXIIIReview for Midterm II 4 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus, f0(x) = ln(π) · f (x) = ln(π)πx.Math 1151, Calculus I Lecture XXIIIReview for Midterm II 5 / 9Compute some derivativesCompute f0(x) where f (x) = πx.STRATEGY 1:f (x) = πx= eln(πx)= ex ·ln(π)Since ln(π) is constant f0(x) =ddx[ex ln(π)] = ln(π)ex ln(π)If you please, this simplifies f0(x) = ln(π) · πxSTRATEGY 2: Logarithmic differentiation.ddx[ln(|f (x)|)] =f0(x)f (x)so that f0(x) = f (x ) ·ddx[ln(|f (x)|)]ddx[ln(|f (x)|)] =ddx[ln(πx)] =ddx[x ln(π)] = ln(π)Thus,