Given: The function y = x²e7. Differentiate equation (1) with respect to x on both sides. - 4(re) dy dx dx Let u = x² and v = e7. Apply product rule and obtain as, *(fe#) =r4 ()+e4() (2) d dx dx dx The derivatives are found as, #(e) = (") (1-x-) = (-r") d dx dx %3D (x-2) 숲 (12) = 2x %3D Thus ouotion (2) hooomor
Given: The function y = x²e7. Differentiate equation (1) with respect to x on both sides. - 4(re) dy dx dx Let u = x² and v = e7. Apply product rule and obtain as, *(fe#) =r4 ()+e4() (2) d dx dx dx The derivatives are found as, #(e) = (") (1-x-) = (-r") d dx dx %3D (x-2) 숲 (12) = 2x %3D Thus ouotion (2) hooomor
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 37E
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I'm confused about what they're doing. It looks like they're using the chain rule on the e^-1/x. Do I need to do that every time the exponent has a variable in it?
Transcribed Image Text:Given:
The function y =x²e7.
Differentiate equation (1) with respect to x on both sides.
dy
dx
dx
Let u = x² and v = e7.
Apply product rule and obtain as,
= x² d
dx
d
+ e
dx
dx
The derivatives are found as,
(-) = 1(*)
= e(-x ')
' (x-²)
dx
dx
숲 (12) %3D 2r
dx
Thue oauetinn (2) hoonmor.
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