Please answer with steps. Assume that (1 + xy)(1 + sin x)(1 + smt/)(ln(z + e)) =
ID: 2860112 • Letter: P
Question
Please answer with steps.
Assume that (1 + xy)(1 + sin x)(1 + smt/)(ln(z + e)) = 1. Confirm that (0, 0) is on this curve. Find dy/dx at (x, y) = (0,0) We say that a pair of functions f(x) and g(x), defined for all real numbers, are intercomposeable if f(g(x)) = g(f(x)). Can you find a pair of intercomposeable functions f(x) and g(x) with f(x) g(x) and neither function equal to the identity? What about a triplet of distinct functions each one intercomposeable with the other two and none equal to the identity? A differential equation is an equation that relates a function and its derivative. For example, f'(x) = f(x) is a differential equation satisfied by f(x) = e^x. We've mentioned this example before in class as well as the similar example that g^(4)(x) = g(x) which has solutions of g(x) = cos x and g(x) - sin x. Can you find a solution to the differential equation f'(x) 2f(x)? Related rates: A gelatinous cube is growing. When it is 125 cubic meters, its side lengths are growing at 3 meters per a day. How fast is the cube's volume growing when it is 125 cubic meters?Explanation / Answer
There are multi problems given in the paper. Only one can be solved here at a time. Accordingly, solution to #8 is give below.
The point (0,0) would be on the curve, if it satisfies the given equation. To do that,find out, simplify the LHS of the equation by plugging in x=0, y=0. It would be ( 1+0) (1+0)(1+0) ln e. { sin 0 equal to 0}
Thus LHS is ln e = 1 { ln e equals 1}. Thus LHS = RHS. The point (0,0) lies on the curve.
To find dy/dx, we can first differentiate the equation w.r.t x, first by considering [(1+xy)(1+sin x) (1+ siny)] as one function and ln(x+e) as another function. Using product rule it would be
[(1+xy)(1+sin x) (1+siny)] d/dx (ln x+e) +ln (x+e) d/dx [(1+xy)(1+sin x) (1+sin y)]=0
Or, ln(x+e) d/dx [(1+xy)(1+sin x) (1+siny)] =- (1+xy)(1+sinx)(1+siny) d/dx (ln x+e) .. {d/dx ln x+e) would be 1/(x+e)}
Hence, ln(x+e) d/dx (1+xy) (1+sinx) (1+ sin y)=-(1+xy)(1+sinx)(1+siny) [1/(x+e) ]. At this stage we can put x=0,y=0 in all the terms which are already differentiated. Accordingly,
ln e d/dx( !+xy)(1+sin x) (1+siny)= -1/e ..... {ln e would be 1}
Hence ,d/dx ( 1+xy)(1+sin x) (1+siny) = -1/e. Next consider (1+xy)(1+sinx) as one function and (1+sin y) as another function and use product rule
(1+xy) (1+sin x) ( cosy dy/dx ) +(1+sin y) d/dx ( 1+xy)(1+sin x) =-1/e. Here again put x=0, y=0 in all the terms already diffentiated, to have,
dy/dx + d/dx (1+xy)(1+sin x)= -1/e
dy/dx + (1+xy)(cosx) + (1+sin x) (y+x dy/dx) = -1/e now put x=0, y=0 to have { cos 0=1 and sin 0=0}
dy/dx +1 =-1/e
dy/dx= -1-1/e
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