g'(:1:)= m, where f stew f(m)=9'1(=v)= 1 1 Step 2 x) = 1 1 Step 3 f'(9(w)) = 1 1 Step 4 m = 1 1 Now, find g'(:1:) by differntiating using the quotient rule. 912:) =1 1 Use the inverse function theorem to find the derivative of g(:z:) : {7%. m) = m where aw) = x) stem ashy1(3): \\ i step 2 at) = [i Step 3 f'(9($)) : '' Step 4 m = i i 1 Now, find g'(:1:) by differntiating g(:r:)_using the power rule, since {/5 : 3:3. g'tm) =1: J Use the inverse function theorem to find the derivative of g(x) = arcsin(x). H g' (2) = f'(g(2)) where f(x) = g-1(x) Step 1 f(x) = 9 1(ac) = Step 2 f'(a) = Step 3 f'(g(x) ) = 1 Step 4 f'(g(x)) Now, state g'(x). g(x)Use the inverse function theorem to find the derivative of g(ac) = arcsin(:c). _ 1 f'(9(m)) r(m>=g-1= S me E me E 1 Step4 mm = S Now, state 9"(17). 9'8\") =: 91$) , where it?) = 9'10?) A person stands a feet from the base of a 225-foot tall building and looks up to the top of the building. An angle of elevation 0 is formed by the lines from the top and bottom of the building to the person, as seen in the figure above. As the person moves closer to or farther from the building (i.e., as a changes), the value of 0 will change. 225 ft do Find the rate of change when x = 213. dx OOC Step 1: Write 0 as a function of x. X Question Help: D Post to forumPart 1 of 2 A 40-foot ladder leans against a tall building. The base of the ladder is 3: feet from the base of the building and the ladder makes an angle of 6 with the ground. A person pulls the bottom of the ladder away from the building, causing the top of the ladder to slide down the wall. Note that distance 3 increases and angle 9 decreases. 40 ft Find the rate of change d when m = 11. 3 Part 1. Write 9 as a function of 3:. 9(3): Assume that at) is a differentiable invertible function with values given on the following table. Find the following values. We: winw: dy Find for the function. Simplify your answer. dx (Note that arcsin , arccos , and arctan are also denoted by sin , cos , and tan , respectively.) y = arcsin(10x) dy dacA. Let f(x) = 16 arccos(In(x)). Then f'(x) B. Let g(x) = -3(arctan(x))". Then g'(x) C. Let h(x) = 5 arcsin(7"). Then h'(x) =A boat is sailing towards a cliff that rises 282 meters above water level. Let 6 be the angle of elevation from the boat to the top of the cliff, and let a: be the distance from the boat to the cliff. Find an equation for 6 in terms of a3. 9 Find the rate of change of 6 with respect to as when the boat is m 2129 meters from the cliff. Round your answer to 3 decimal places. Select an answer v A boat is sailing towards a cliff that rises 282 meters above water level. Let 6 be the angle of elevation from the boat to the top of the cliff, and let a: be the distance from the boat to the cliff. Find an equation for 6 in terms of a3. 9 Find the rate of change of 6 with respect to as when the boat is m 2129 meters from the cliff. Round your answer to 3 decimal places. Select an answer v \fFind the slope of the tangent line to the curve 2:v2 2mg + 2y3 2 18 at the point (3, 3). l: If 3:132 + 33: + my = 5 and y(5) = 17, find y'(5) by implicit differentiation. :] Question Helo: D Post to forum Use implicit differentiation to find the equation of the tangent line to the curve my3 + my = 2 at the point (1, 1). The equation of this tangent line can be written in the form y 2 ms: + b where m is: and where b EC] Question Help: D Post to forum