From first principles find the derivative of sqrt(1+x) ?
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QUESTION 2 (a) By the first principles of differentiation, find the following: (i) Derivative of F(x)= F'(-3) 1-X 2 + x (ii)
answer all 4 questions 2+1) π i Given that :-., and arg( 4. Find Find 2. Use the first principles to find the derivative of [20] cos(x-4x) . rt k! 3. Use derivative to show that [10] 4. Find dry, if y = xx, 2+1) π i Given that :-., and arg( 4. Find Find 2. Use the first principles to find the derivative of [20] cos(x-4x) . rt k! 3. Use derivative to show that [10] 4. Find dry,...
(1 point) The derivative of f(x,y) at (6,5) in the direction 1/sqrt(53) <−7,−2> is 8 and the derivative of f(x,y) at (6,5) in the direction 1/sqrt(17)<−4,1> is −5. What is the derivative of f(x,y) at the point (6,5) in the direction 1/sqrt(89)<8,−5>? Please show all steps.
Find f(x) using first principles where f(x)=(x^3+1)/X^2 fferentiate from first principles the function f(x) = +1 upute the consumer surplus at prica
PLEASE FIND THE DERIVATIVE FOR y= 1./sqrt(f) + 2.0*log10((rough/d)/3.7 + 2.51./(Re.*sqrt(f)));
1. find the derivative of f(×,y)=-4yx^3+xy^2 at P(1,1) in forward direction set by the line r(t)=(1+sqrt(2)t+sqrt(2)t 2. find an equation for the tangent plain at point P x^3+y^3=3xyz P(2,1,3/2)
Find the general solution of following equation x' = [ -3 sqrt(2) sqrt(2) -2 ] *x It's a matrix problem, so x' = [ ] x Hopefully that makes sense, thanks in advance
1) Find the directional derivative of the function at the given point and in the direction of the vector u as shown when f(x,y)= sen(2x+3y); (-6,4); u=(1/2)(sqrt(3)),-1) POSSIBLE ANSWERS A) sqrt(3)-(3/2) B) (3/2)+sqrt(3) C) (3/2)-sqrt(3) D) -(3/2)-sqrt(3) 2) Find the direction in which the function is growing or decreasing more rapidly at the point shown: f(x,y)=x(e^y)-lnx; (4,0) POSSIBLE ANSWERS: A) u=(3/(sqrt(265)) , 16/(sqrt(265))) B)u=(3/(sqrt(265)) , -16/(sqrt(265))) C)u=(16/(sqrt(265)) , 3/(sqrt(265))) D)u=(-3/(sqrt(265)) , 16/(sqrt(265)))
find the derivative 6x<+2 ln(x), (1 point) Find the derivative with respect to x of h(x) = h'(x) =
The first derivative of a cont ys f(X) is y' = x(x-6r Find y" and then use the graphing procedure to sketch the general shape of the graph of t The first derivative of a cont ys f(X) is y' = x(x-6r Find y" and then use the graphing procedure to sketch the general shape of the graph of t