Part 1: Let 2sinx+7cosy=2. dydx= d2ydx2= Part 2: Four functions are given below.

Question

Part 1:

Let 2sinx+7cosy=2.

dydx=

d2ydx2=

Part 2:

Four functions are given below. Perform the indicated compositions to determine which functions are inverse to each other. Be sure to simplify the results.

f(x)g(x)h(x)j(x)====17x+15x1715x17151717x+255

f(g(x))=    g(f(x))=
Conclusion: f and g  ? are are not  inverses.

f(h(x))=    h(f(x))=
Conclusion: f and h  ? are are not  inverses.

j(g(x))=    g(j(x))=
Conclusion: g and j  ? are are not  inverses.

Part 3:

Determine whether the function below is its own inverse.

f(x)=3x24x3,x34.

f(f(x))=
Conclusion: f  ? is is not  the inverse of itself.

Part 4:

An invertible function f(x) is given along with a point that lies on its graph. Using Theorem 22 (p. 116 of APEX Calculus v.3), evaluate (f1)(x) at the indicated value.

f(x)=x+sin3x,6x6Point=(18,0.674533)

(f1)(0.674533)=

Explanation / Answer

1)

2sinx+7cosy=2.

differentiate with respect to x

2cosx -7sinydy/dx =0

=>dy/dx =2cosx/(7siny)

differentiate with respect to x , quotient rule:(u/v)'=(u'v -uv')/v2

d2y/dx2=[(-2sinx)(7siny) -(2cosx)(7cosy)dy/dx ]/(7siny)2

we have dy/dx =2cosx/(7siny)

d2y/dx2=[(-2sinx)(7siny) -(2cosx)(7cosy)(2cosx/(7siny)) ]/(7siny)2

d2y/dx2=[(-14sinxsiny) -(4cos2x coty) ]/(7siny)2

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