Hi there
Please help me solve these problems. I tried, but I can't figure them out.
1) If is a differentiable function of then the slope of the tangent to the curve at the point where [tex]y=1[tex] is..?
2)Find the point on the graph of y=x^(1/2) between (1,1) and (9,3) at whtich the tangent to the graph has the same slop as the line through (1,1) and (9,3)
3) Let f and g be differential functions such that :
f(1) = 4
g(1) = 3
f'(3) = -5
f'(1)= -4
g'(1)= -3
g'(3) = 2
If h(x) = f(g(x)), then h'(1) =
4) Find the derivative and the equation of the tangent line to
5) Find the second derivative of
6) Use the tangent line approximation of y= square root of x at x=16 to find the approximate value of the square root of 17
And also, what does differentiable mean? How can you tell on a graph if a point is differentiable or not?
Thanks for the help
find the slope of the line that passes through (1,1) and (9,3) (do you know how to do that?)
then, find the derivative of y = x^(1/2) and set it equal to the value you found above. solve for x, this will give you the x-value of the point you are after. then plug this x-value into the original function to solve for the corresponding y-value, and then you'll have your point
remember, when doing implicit differentiation, we keep track of what we are differentiating and what we are differentiating with respect to. here we are differentiating with repsect to x. so when we differentiate an x-term, we attach dx/dx to it (since we differentiated an x-term with respect to x). but derivative notations can function as fractions and this cancels and we get 1, so we don't write it. when we differentiate a y-term, we attach dy/dx (since we differentiated a y-term with respect to x). we justify doing this by the chain rule. you can look it up in your text for more info.
i will use y' to mean dy/dx to save on typing
differentiating implicitly we get:
...........note that i used the chain rule for the first term
now, solve for y' and plug in the required x and y-values
finding the derivative is very similar to the first question.
to find the tangent line, we need to know at what point. when you have the point, use the point-slope form:
where is the slope of the tangent line at the point and is a point the line passes through
just solve for y
this is the formula for the linear approximation of a function.
here is the value for which we want to find for, but we can't because it does not have a nice answer, and is a point close to for which we do know the exact value of the function.
here , and
thus,
now continue
"differentiable" means we can differentiate it. that is, the function's derivative can be found, or exists at a certain point, or on some interval.
recall how you defined the derivative. we started out with the slope formula for the equation of a line. thus, in layman's terms, a function is differentiable at a point, if it is approximately linear at that point. that is, if we zoom in close to a point, the function looks like a straight line. what this translates to is the following.How can you tell on a graph if a point is differentiable or not?
- we cannot find the derivative at points where a function takes a sharp turn, e.q. y = |x| at x = 0
- we cannot find the derivative of a function where it is discontinuous