calc

[xtrmk]

can someone help me on these

1) find y'' if y = sqrt(x^2+16)
.: i rewrote it (x^2+16)^1/2
.: y' = 1/2(x^2+16)^-(1/2) (2x)

now what? do i simplify then take the derivative again?

2) If y = sin^3 (1-2x) find dy/dx
.: I don't know how to start. Do I just use cos^3(1-2x) or something..

3) Given just a graph, how do i find the instantaneous rate of change at a specific x value? I am given nothing on the graph except the x and y values and a graph

thnx in advance..

 

[daon]

1) you got it
2) Rewrite that as [Sin(1-2x)]³. Then you have three composite functions: f(x)=x³, g(x)=Sin(x), h(x)=1-2x, and you have (f o g o h)(x) = f(g(h(x))).

Use chain rule: If y=f(g(h(x))), then dy = f'(g(h(x)))*g'(h(x))*h'(x)dx.

3) The instantaneous rate of change is when the secant is drawn between two infinitely close points on the graph. You can draw a line tangent to the graph at any point to determine an approximate instantaneous rate of change.

Hope that helps,
-Daon