How do I find the derivative on an inverse graph geometrically?
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
819742
The first derivative is the slope of the line that is tangent to the graph.  If there is a maximum or minimum, the tangent line at that point will be horizontal, so its slope is 0.  \n\nThe second derivative is 0 when there is an inflection point. An inflection point is a place on the graph where it changes the direction of the concavity. \n\nIn the two cases above, the converse is not necessarily true.  That is, just because the first derivative is 0 doesn't mean there has to be a max/min there.\n\n If the second derivative is positive, the graph is concave upward which means like a regular "U".  If it's negative, it's concave downward like an upside-down "U".
669809
Start by finding the partial derivatives.
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
208411
Well, first you get the point where the line is tangent to your function, then compute the derivative of the function.  Plug in your x-coordinate to the derivative and the y-coordinate is the slope at the given point.\n\nfor example: y=3x^2 point=(2,1) (taking a random point here)\n             y`=6x\n             slope of tangent line at (2,1) would be 12 (6 x 2)
945033
Derivative is y=cos x -5. Substituting in 0 gives you the slope -4. Now plug in the coordinates (0,3) to get the line, which is y=-4x+3
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
758976
The total differential is [∂F/∂x]dx +[ ∂F/∂y]dy\n\n∂F/∂x = 4x^3*y^-1\n∂F/∂y = x^4*(-y^-2) = -x^4y^-2
17334
Geometric -- multiplying the previous term by 4
883736
x=-t/2\n2x=-t\n-2x=t\ny=t^2\ny=(-2x)^2\ny=4x^2\nThe graph is a parabola with vertex at (0,0), and in the x-interval [0,5], the end falls at (5, 100).
1101719
Step 1: Find the slope of the line x-2y = 2.  Call it m. \n\nStep 2: Find the derivative f'(x) of the function f(x) = (x-1)/(x+1).\n\nStep 3: Find all values of x where f'(x) = m.\n\nStep 4: For each such value, find its corresponding y-coordinate by setting y=f(x), and then find the equation of the line that goes through that point with slope m.\n\nGood luck!
878479
To find the inverse function switch the x's and y's\n\ny=x^2-3\nbecomes\nx=y^2-3\nand after simplifying\ny=+/-sqrt(x+3) -- don't forget the plus or minus, you need both for this one!
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
1015799
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
609321
One is the inverse of the other
1068232
Use quotient rule to get the derivative, which, simplified, is:\n\n-30*(2x+3)/((3x^2+9x+7)^2)\n\nSet it equal to zero and solve.  Since the denominator cannot be zero, set the numerator equal to zero\n\n2x+3=0\nx=(-3/2)\n\nI checked it, and it indeed is the solution. \n\nBy the way, you don't have to worry about an undefined points of this function making problems in the graph or in the max/min values in the derivative.  The denominator has imaginary roots.
373083
Well, I'm not going to answer the question for you, but I'll tell you how to do it.\n\nWhen you're graphing the function, you set up a table with f(x) and x as the two columns, then solve for x for any set of numbers, like -3, -2, -1, 0, 1, 2, 3, and that's probably more than enough. Graph those on a piece of graph paper (f(x)=y, and then x). \n\nNow, to find the INVERSE, you simple switch the X and the Y. Instead of solving for x, you solve for f(x). So, your problem would be written x= -2/3y + 5. An easier way to do it, take the table you made before, and switch the columns, so that the solutions for x are underneath f(x) and the numbers you picked for f(x) are underneath the x column. Graph those coordinates. When you look at the graph, the inverse should be a mirror reflection of the first one, with  the graph of y=x as the line of symmetry. \n\nThat's how you find the inverse of any problem. Sorry if anything's unclear, if you need a better visual, I'm sure you could google up this exact method.