dec 15 study set

Which of the following requires the use of implicit differentiation to find dydx ?

y=ln(y+x)+x2y=ln(y+x)+x2 Correct. In this expression yy is an implicitly-defined function and the independent variable xx cannot be isolated. Therefore implicit differentiation is needed to find the derivative dydxdydx.

For which of the following functions would the quotient rule be considered the best method for finding the derivative?

y=sin(2x+1)2x+1y=sin(2x+1)2x+1

If y=2lnx, then d4ydx4=

−12x4−12x4

If sin(x+y)=3x−2y, then dydx=

Correct. The chain rule is the basis for implicit differentiation as well as the differentiation of a composite function. cos(x+y)(1+dydx)=3−2dydxcos(x+y)(1+dydx)=3−2dydx cos(x+y)+cos(x+y)dydx=3−2dydxcos(x+y)+cos(x+y)dydx=3−2dydx (2+cos(x+y))dydx=3−cos(x+y)(2+cos(x+y))dydx=3−cos(x+y) dydx=3−cos(x+y)2+cos(x+y)

The table above gives selected values for a differentiable and decreasing function f and its derivative. If g(x)=f−1(x) for all x, which of the following is a correct expression for g′(2) ?

Correct. This value can be confirmed using the chain rule and the definition of an inverse function. Since f(g(x))=xf(g(x))=x, it follows that ddxf(g(x))=f′(g(x))g′(x)=ddx(x)=1⇒g′(x)=1f′(g(x))ddxf(g(x))=f′(g(x))g′(x)=ddx(x)=1⇒g′(x)=1f′(g(x)). Therefore, g′(2)=1f′(g(2))=1f′(4)=−15g′(2)=1f′(g(2))=1f′(4)=−15.

Which of the following methods can be used to find the derivative of y=arcsinx with respect to x ?

2 and 3. Correct. The definition of the inverse sine function means that sin(arcsinx)=xsin(arcsin⁡x)=x. Using method II, ddxsin(arcsinx)=cos(arcsinx)⋅(ddxarcsinx)=ddx(x)=1⇒ddxarcsinx=1cos(arcsinx)ddxsin(arcsin⁡x)=cos(arcsin⁡x)⋅(ddxarcsin⁡x)=ddx(x)=1⇒ddxarcsin⁡x=1cos(arcsin⁡x). If θ=arcsinxθ=arcsin⁡x, then −π2≤θ≤π2−π2≤θ≤π2 and sinθ=xsin⁡θ=x. Therefore, cosθcos⁡θ is nonnegative and cos2θ=1−sin2θ=1−x2⇒cosθ=1−x2−−−−−√cos2θ=1−sin2θ=1−x2⇒cos⁡θ=1−x2. The derivative is then ddxarcsinx=1cos(arcsinx)=1cosθ=11−x2√ddxarcsin⁡x=1cos(arcsin⁡x)=1cos⁡θ=11−x2 for −1<x<1−1<x<1. Using method III, cosydydx=1⇒dydx=1cosycos⁡ydydx=1⇒dydx=1cos⁡y. Since y=arcsinxy=arcsin⁡x implies that −π2≤y≤π2−π2≤y≤π2, the value of cosycos⁡y is nonnegative. Then, cos2y=1−sin2y=1−x2⇒cosy=1−x2−−−−−√cos2y=1−sin2y=1−x2⇒cos⁡y=1−x2. Therefore, ddx(arcsinx)=dydx=1cosy=11−x2√ddx(arcsin⁡x)=dydx=1cos⁡y=11−x2 for −1<x<1−1<x<1. Method I cannot be used since 1sinx1sin⁡x is not equal to arcsinxarcsin⁡x, the inverse sine function.

The figure above shows the graph of f′, the derivative of the function f. At which of the four indicated values of x is f′′(x) greatest?

Answer B Correct. The second derivative f′′f″ is the rate of change of the first derivative f′f′. The value of f′′(x)f″(x) therefore represents the slope of the line tangent to the graph of f′f′ at xx. The tangent lines at x=Ax=A and x=Bx=B both have positive slopes, but the tangent line at x=Bx=B is steeper than the tangent line at x=Ax=A. Therefore f′′(B)f″(B) is greater than f′′(A)f″(A). The tangent line at x=Cx=C is horizontal, so f′′(C)=0f″(C)=0. The tangent line at x=Dx=D has a negative slope, so f′′(D)<0f″(D)<0. Therefore f′′(B)f″(B) has the greatest value.

Let y=f(x) be a twice-differentiable function such that f(1)=3 and dydx=4y2+7x2−−−−−−−√. What is the value of d2ydx2 at x=1 ?

Answer C Correct. The second derivative can be found using implicit differentiation of the first derivative, then evaluated at the point (1,3)(1,3). d2ydx2=ddx(dydx)=ddx(4y2+7x2−−−−−−−√)=2⋅2ydydx+14xy2+7x2√d2ydx2=ddx(dydx)=ddx(4y2+7x2)=2⋅2ydydx+14xy2+7x2 Since dydx∣∣(1,3)=432+7−−−−−√=416−−√=16dydx|(1,3)=432+7=416=16, d2ydx2=2⋅2ydydx+14xy2+7x2√∣∣∣(1,3)=2⋅2(3)(16)+144=55d2ydx2=2⋅2ydydx+14xy2+7x2|(1,3)=2⋅2(3)(16)+144=55.

For which of the following functions is the chain rule an appropriate method to find the derivative with respect to x ? y=sin(3x2) y=extanx y=18x4−2x

Correct. Each of the functions in I and III is a composition of component functions. In I, let f(x)=sinxf(x)=sin⁡x and let g(x)=3x2g(x)=3x2. Then sin(3x2)=f(g(x))sin(3x2)=f(g(x)), and the chain rule would give ddxsin(3x2)=f′(g(x))g′(x)=cos(3x2)6xddxsin(3x2)=f′(g(x))g′(x)=cos(3x2)6x. In III, let f(x)=1x=x−1f(x)=1x=x−1 and g(x)=8x4−2xg(x)=8x4−2x. Then 18x4−2x=f(g(x))18x4−2x=f(g(x)), and the chain rule would give ddx18x4−2x=f′(g(x))g′(x)=−1(8x4−2x)2⋅(32x3−2)ddx18x4−2x=f′(g(x))g′(x)=−1(8x4−2x)2⋅(32x3−2). In II, the function y=extanxy=extan⁡x is the product of the exponential function and the tangent function and is not the composition of two component functions. The product rule would be used to find the derivative, and the chain rule would not be appropriate.

ddx(cos−1x)=

Correct. If y=cos−1xy=cos⁡−1x, then cosy=xcos⁡y=x. Using the chain rule to differentiate this implicit relation gives −sinydydx=1⇒dydx=−1siny−sin⁡ydydx=1⇒dydx=−1sin⁡y. Since y=cos−1xy=cos⁡−1x implies that 0≤y≤π0≤y≤π, the value of sinysin⁡y is nonnegative. Then sin2y=1−cos2y=1−x2⇒siny=1−x2−−−−−√sin⁡2y=1−cos⁡2y=1−x2⇒sin⁡y=1−x2. Therefore, dydx=−1siny=−11−x2√dydx=−1sin⁡y=−11−x2.

ddx(sin−1x)∣∣x=12=

Correct. If y=sin−1xy=sin⁡−1x, then siny=xsin⁡y=x. Using the chain rule to differentiate this implicit relation gives cosydydx=1⇒dydx=1cosycos⁡ydydx=1⇒dydx=1cos⁡y. Since y=sin−1xy=sin⁡−1x implies that −π2≤y≤π2−π2≤y≤π2, the value of cosycos⁡y is nonnegative. Then cos2y=1−sin2y=1−x2⇒cosy=1−x2−−−−−√cos⁡2y=1−sin⁡2y=1−x2⇒cos⁡y=1−x2. Therefore, dydx=1cosy=11−x2√dydx=1cos⁡y=11−x2. This is evaluated at x=12x=12.

The point (−2,4) lies on the curve in the xy-plane given by the equation f(x)g(y)=17−x−y, where f is a differentiable function of x and g is a differentiable function of y. Selected values of f, f′, g, and g′ are given in the table above. What is the value of dydx at the point (−2,4) ?

Correct. The chain rule is the basis for implicit differentiation. f′(x)g(y)+f(x)g′(y)dydx=−1−dydxf′(x)g(y)+f(x)g′(y)dydx=−1−dydx (1+f(x)g′(y))dydx=−1−f′(x)g(y)⇒dydx=−1−f′(x)g(y)1+f(x)g′(y)(1+f(x)g′(y))dydx=−1−f′(x)g(y)⇒dydx=−1−f′(x)g(y)1+f(x)g′(y) dydx∣∣(−2,4)=−1−f′(−2)g(4)1+f(−2)g′(4)=−1−201+6=−217=−3

If g(x)=lnx and f is a differentiable function of x, which of the following is equivalent to the derivative of f(g(x)) with respect to x ?

Correct. The chain rule provides a way to differentiate a composition of functions. The "outside" function is f(x)f(x) and the "inside" function is g(x)=lnxg(x)=ln⁡x. Then g′(x)=1xg′(x)=1x and the chain rule can be used, as follows. ddxf(lnx)=ddxf(g(x))=f′(g(x))⋅g′(x)=f′(lnx)⋅1x

What is the slope of the line tangent to the curve y3−xy2+x3=5 at the point (1,2) ?

Correct. The point (1,2)(1,2) is on the curve since x=1x=1, y=2y=2 satisfies the implicit equation of the curve. The chain rule is the basis for implicit differentiation. 3y2dydx−y2−2xydydx+3x2=03y2dydx−y2−2xydydx+3x2=0 At the point (1,2)(1,2), 12dydx−4−4dydx+3=0⇒8dydx−1=0⇒dydx=1812dydx−4−4dydx+3=0⇒8dydx−1=0⇒dydx=18.

Let f be the increasing function defined by f(x)=x3+2x2+4x+5, where f(−1)=2. If g is the inverse function of f, which of the following is a correct expression for g′(2) ?

Correct. This can be confirmed using the chain rule and the definition of an inverse function. Since f(g(x))=xf(g(x))=x, it follows that ddxf(g(x))=f′(g(x))g′(x)=ddx(x)=1⇒g′(x)=1f′(g(x))ddxf(g(x))=f′(g(x))g′(x)=ddx(x)=1⇒g′(x)=1f′(g(x)). Therefore, g′(2)=1f′(g(2))=1f′(−1)g′(2)=1f′(g(2))=1f′(−1).

Which of the following expressions can be differentiated using the product rule

Correct. This expression is the product of x2x2 and tan−1xtan⁡−1x, and therefore requires use of the product rule to differentiate the expression.

Let f be a differentiable function. If h(x)=(1+f(3x))2, which of the following gives a correct process for finding h′(x) ?

Correct. This requires using the chain rule twice to compute the derivative. h′(x)=2(1+f(3x))⋅(ddx(1+f(3x)))=2(1+f(3x))(f′(3x)⋅ddx(3x))=2(1+f(3x))⋅f′(3x)⋅3

The graph of the decreasing differentiable function f is shown above. Also shown is the line tangent to the graph of f at the point (4,6). Let g be the inverse of f. Which of the following statements about g′ is true?

Correct. This statement can be confirmed using the chain rule and the definition of an inverse function. The line tangent to the graph of ff at the point (4,6)(4,6) has slope f′(4)=−34f′(4)=−34. Since g(x)=f−1(x)g(x)=f−1(x), then g(6)=4g(6)=4 and f(g(x))=xf(g(x))=x. By the chain rule, ddxf(g(x))=f′(g(x))g′(x)=ddx(x)=1⇒g′(6)=1f′(g(6))=1f′(4)=1−34=−43