LESSON 8: INTRODUCTION TO FUNCTIONS OF SEVERAL VARIABLES MATH 6020 FALL 208 ELLEN WELD. Partial Derivatives We aress how to take a erivative of a function of several variables. Although we won t get into the etails, the iea is that we take a erivative with respect to a irection. What we mean is this: if we have a function of several variables, we choose variable an take the erivatives thinking of all the other variables as s. But this type of erivative oesn t give the entire picture of what the function is oing so we call these partial erivatives. Ex. Let f(x, y) x + 2y an suppose we want to fin its partial erivatives. First, we nee to choose a variable, say x. Secon, we think of the other variables (in this case just y) as with respect to x. This means we think of x an y as acting totally inepenently so x changing oesn t affect y. We use a special notation enote this concept: x. So the partial erivative with respect to (wrt) x is: f(x, y) (x + 2y) x x x (x) + x (2y) + 0. y oes not change wrt to x Notice that x (x) (x). This is because with respect to x, x erivative as we ve always one it. x is exactly the We use the same line of thinking when we take y an hol x fixe. Here, x is a with respect to y. Again, we have our own notation: y. The partial erivative with respect to y is: y f(x, y) (x + 2y) y y (x) + (2y) 0 + 2 2. y x oes not change wrt y Again, we see that y (2y) (2y) because with respect to y, y erivative as before. y is the same y
2 ELLEN WELD Remark. This is not a, an we will call it el. is use exclusively for partial erivatives. Note 2. We will use a variety of notation for partial erivatives but they will mean the same thing. For example, if our function is z f(x, y), we can write an f x f x (x, y) f f(x, y) x x z x f y f y (x, y) f f(x, y) y y z y. For f(x, y) x + 2y, our partial erivatives are f x an f y 2. Now is the time to review all of the ifferentiation rules you have forgotten. Examples.. Fin f x, f y if f(x, y) e x2 + ln y 2. Solution: f x (x, y) x (ex2 + ln y 2 ) x (ex2 ) + x (ln y2 ) 2xe x2 0 f y (x, y) y (ex2 + ln y 2 ) y (ex2 ) + y (ln y2 ) 2y y 2 2 y 0 f x 2xe x2 an f y 2 y 2. Fin f x, f y if f(x, y) y cos x. Solution: For this problem, we want to remember that whenever c was a x (cx3 ) c x (x3 ). We have the same property for partial erivatives: x (cx3 ) c x (x3 ). Even more, any function of y is with respect to x. So x ((sin y)x3 ) sin(y) x (x3 ).
AN UNOFFICIAL GUIDE TO MATH 6020 FALL 208 3 With this in min, we compute f x (x, y) (y cos x) x y f y (x, y) y (y cos x) cos x wrt y (cos x) y sin x x y (y) cos x f x y sin x an f y cos x. 3. Fin f x (, 0) an f y (, 0) if f(x, y) 3x y y. Solution: The ifference between this example an the examples above is that here we nee to ifferentiate an then evaluate the erivative at the point (, 0). Differentiating with respect to x, ( ) 3x y y f x (x, y) x Hence, y 3 (3x y) (3 0) x y y. f x (, 0) 3 0 3. Now, to ifferentiation with respect to y, we will nee to use the quotient rule (we coul also use the prouct rule after a small rewrite). So f y (x, y) ( ) 3x y y y ( y) y (3x y) (3x y) ( y) y ( y) 2 ( y)(0 ) (3x y)( ) ( y) 2 y y + 3x ( y) 2 3x ( y) 2. f y (, 0) 3() ( 0) 2 2.
4 ELLEN WELD Finally, f x (, 0) 3 an f y (, 0) 2. 4. Fin f x, f y if f(x, y) e x2y. Solution: Recall that if we were consiering a function of a single variable, say e 3x+x2, its erivative with respect to x is ( ) x e3x+x2 x (3x + x2 ) e 3x+x2 (3 + 2x)e 3x+x2 by the chain rule. The chain rule still applies for partial erivatives. f x x (ex2y ) x (x2 y)e x2y y x (x2 )e x2y y(2x)e x2y 2xye x2 y f y y (ex2y ) y (x2 y)e x2y x 2 y (y)ex2y x 2 e x2 y wrt y So, f x 2xye x2y an f y x 2 e x2y. 5. Fin f x (, 0) if f(x, y) ln(ln(y)x). Solution: Recall that x ln(g(x)) g (x) g(x). This rule still follows for partial erivatives: f x (x, y) x (ln(ln(y)x)) x (ln(y)x) ln(y)x f x (, 0). ln(y) x (x) ln(y)x ln(y) ln(y)x x 6. Fin f x, f y if f(x, y) xy sin(xy). Solution: We nee to use the prouct rule an chain rule. f x (x, y) (xy sin(xy)) x xy x (sin(xy)) + (xy) sin(xy) x xy( y cos(xy)) + y sin(xy) x (xy)
AN UNOFFICIAL GUIDE TO MATH 6020 FALL 208 5 Thus xy 2 cos(xy) + y sin(xy) f y (x, y) xy y (sin(xy)) + (xy) sin(xy) y xy( x y (xy) cos(xy)) + x sin(xy) x 2 y cos(xy) + x sin(xy) f x xy 2 cos(xy) + y sin(xy) an f y x 2 y cos(xy) + x sin(xy).