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Key concepts

  • A partial derivative is a derivative involving a function of more than one independent variable.
  • To calculate a partial derivative with respect to a given variable, treat all the other variables as constants and use the usual differentiation rules.
  • Higher-order partial derivatives can be calculated in the same way as higher-order derivatives.

Key equations

  • Partial derivative of f with respect to x
    f x = lim h 0 f ( x + h , y ) f ( x , y ) h
  • Partial derivative of f with respect to y
    f y = lim k 0 f ( x , y + k ) f ( x , y ) k

For the following exercises, calculate the partial derivative using the limit definitions only.

z x for z = x 2 3 x y + y 2

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z y for z = x 2 3 x y + y 2

z y = −3 x + 2 y

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For the following exercises, calculate the sign of the partial derivative using the graph of the surface.

A partial paraboloid with vertex at the origin and pointing up.

f x ( −1 , 1 )

The sign is negative.

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f x ( 0 , 0 )

The partial derivative is zero at the origin.

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For the following exercises, calculate the partial derivatives.

z x for z = sin ( 3 x ) cos ( 3 y )

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z y for z = sin ( 3 x ) cos ( 3 y )

z y = −3 sin ( 3 x ) sin ( 3 y )

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z x and z y for z = x 8 e 3 y

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z x and z y for z = ln ( x 6 + y 4 )

z x = 6 x 5 x 6 + y 4 ; z y = 4 y 3 x 6 + y 4

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Find f y ( x , y ) for f ( x , y ) = e x y cos ( x ) sin ( y ) .

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Let z = e x y . Find z x and z y .

z x = y e x y ; z y = x e x y

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Let z = ln ( x y ) . Find z x and z y .

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Let z = tan ( 2 x y ) . Find z x and z y .

z x = 2 sec 2 ( 2 x y ) , z y = sec 2 ( 2 x y )

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Let z = sinh ( 2 x + 3 y ) . Find z x and z y .

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Let f ( x , y ) = arctan ( y x ) . Evaluate f x ( 2 , −2 ) and f y ( 2 , −2 ) .

f x ( 2 , −2 ) = 1 4 = f y ( 2 , −2 )

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Let f ( x , y ) = x y x y . Find f x ( 2 , −2 ) and f y ( 2 , −2 ) .

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Evaluate the partial derivatives at point P ( 0 , 1 ) .

Find z x at ( 0 , 1 ) for z = e x cos ( y ) .

z x = cos ( 1 )

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Given f ( x , y , z ) = x 3 y z 2 , find 2 f x y and f z ( 1 , 1 , 1 ) .

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Given f ( x , y , z ) = 2 sin ( x + y ) , find f x ( 0 , π 2 , −4 ) , f y ( 0 , π 2 , −4 ) , and f z ( 0 , π 2 , −4 ) .

f x = 0 , f y = 0 , f z = 0

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The area of a parallelogram with adjacent side lengths that are a and b , and in which the angle between these two sides is θ , is given by the function A ( a , b , θ ) = b a sin ( θ ) . Find the rate of change of the area of the parallelogram with respect to the following:

  1. Side a
  2. Side b
  3. Angle θ
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Express the volume of a right circular cylinder as a function of two variables:

  1. its radius r and its height h .
  2. Show that the rate of change of the volume of the cylinder with respect to its radius is the product of its circumference multiplied by its height.
  3. Show that the rate of change of the volume of the cylinder with respect to its height is equal to the area of the circular base.

a. V ( r , h ) = π r 2 h b. V r = 2 π r h c. V h = π r 2

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Calculate w z for w = z sin ( x y 2 + 2 z ) .

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Find the indicated higher-order partial derivatives.

f x y for z = ln ( x y )

f x y = 1 ( x y ) 2

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f y x for z = ln ( x y )

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Let z = x 2 + 3 x y + 2 y 2 . Find 2 z x 2 and 2 z y 2 .

2 z x 2 = 2 , 2 z y 2 = 4

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Given z = e x tan y , find 2 z x y and 2 z y x .

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Given f ( x , y , z ) = x y z , find f x y y , f y x y , and f y y x .

f x y y = f y x y = f y y x = 0

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Given f ( x , y , z ) = e −2 x sin ( z 2 y ) , show that f x y y = f y x y .

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Show that z = 1 2 ( e y e y ) sin x is a solution of the differential equation 2 z x 2 + 2 z y 2 = 0 .

d 2 z d x 2 = 1 2 ( e y e y ) sin x d 2 z d y 2 = 1 2 ( e y e y ) sin x d 2 z d x 2 + d 2 z d y 2 = 0

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Find f x x ( x , y ) for f ( x , y ) = 4 x 2 y + y 2 2 x .

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Let f ( x , y , z ) = x 2 y 3 z 3 x y 2 z 3 + 5 x 2 z y 3 z . Find f x y z .

f x y z = 6 y 2 x 18 y z 2

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Let F ( x , y , z ) = x 3 y z 2 2 x 2 y z + 3 x z 2 y 3 z . Find F x y z .

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Given f ( x , y ) = x 2 + x 3 x y + y 3 5 , find all points at which f x = f y = 0 simultaneously.

( 1 4 , 1 2 ) , ( 1 , 1 )

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Given f ( x , y ) = 2 x 2 + 2 x y + y 2 + 2 x 3 , find all points at which f x = 0 and f y = 0 simultaneously.

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Given f ( x , y ) = y 3 3 y x 2 3 y 2 3 x 2 + 1 , find all points on f at which f x = f y = 0 simultaneously.

( 0 , 0 ) , ( 0 , 2 ) , ( 3 , −1 ) , ( 3 , −1 )

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Given f ( x , y ) = 15 x 3 3 x y + 15 y 3 , find all points at which f x ( x , y ) = f y ( x , y ) = 0 simultaneously.

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Show that z = e x sin y satisfies the equation 2 z x 2 + 2 z y 2 = 0 .

2 z x 2 + 2 z y 2 = e x sin ( y ) e x sin y = 0

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Show that f ( x , y ) = ln ( x 2 + y 2 ) solves Laplace’s equation 2 z x 2 + 2 z y 2 = 0 .

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Show that z = e t cos ( x c ) satisfies the heat equation z t = e t cos ( x c ) .

c 2 2 z x 2 = e t cos ( x c )

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Find lim Δ x 0 f ( x + Δ x ) f ( x , y ) Δ x for f ( x , y ) = −7 x 2 x y + 7 y .

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Find lim Δ y 0 f ( x , y + Δ y ) f ( x , y ) Δ y for f ( x , y ) = −7 x 2 x y + 7 y .

f y = −2 x + 7

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Find lim Δ x 0 Δ f Δ x = lim Δ x 0 f ( x + Δ x , y ) f ( x , y ) Δ x for f ( x , y ) = x 2 y 2 + x y + y .

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Find lim Δ x 0 Δ f Δ x = lim Δ x 0 f ( x + Δ x , y ) f ( x , y ) Δ x for f ( x , y ) = sin ( x y ) .

f x = y cos x y

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The function P ( T , V ) = n R T V gives the pressure at a point in a gas as a function of temperature T and volume V . The letters n and R are constants. Find P V and P T , and explain what these quantities represent.

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The equation for heat flow in the x y -plane is f t = 2 f x 2 + 2 f y 2 . Show that f ( x , y , t ) = e −2 t sin x sin y is a solution.

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The basic wave equation is f t t = f x x . Verify that f ( x , t ) = sin ( x + t ) and f ( x , t ) = sin ( x t ) are solutions.

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The law of cosines can be thought of as a function of three variables. Let x , y , and θ be two sides of any triangle where the angle θ is the included angle between the two sides. Then, F ( x , y , θ ) = x 2 + y 2 2 x y cos θ gives the square of the third side of the triangle. Find F θ and F x when x = 2 , y = 3 , and θ = π 6 .

F θ = 6 , F x = 4 3 3

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Suppose the sides of a rectangle are changing with respect to time. The first side is changing at a rate of 2 in./sec whereas the second side is changing at the rate of 4 in/sec. How fast is the diagonal of the rectangle changing when the first side measures 16 in. and the second side measures 20 in.? (Round answer to three decimal places.)

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A Cobb-Douglas production function is f ( x , y ) = 200 x 0.7 y 0.3 , where x and y represent the amount of labor and capital available. Let x = 500 and y = 1000 . Find δ f δ x and δ f δ y at these values, which represent the marginal productivity of labor and capital, respectively.

δ f δ x at ( 500 , 1000 ) = 172.36 , δ f δ y at ( 500 , 1000 ) = 36.93

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The apparent temperature index is a measure of how the temperature feels, and it is based on two variables: h , which is relative humidity, and t , which is the air temperature.

A = 0.885 t 22.4 h + 1.20 t h 0.544 . Find A t and A h when t = 20 ° F and h = 0.90 .

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Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Practice Key Terms 4

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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