Top answers

Differentiate w.r.t x the expression arccos(x).

Using implicit differentiation, let y equal arccos(x) : y=arccos(x). So x = cos(y), and dx/dy = -sin(y). dy/dx is therefore -1/sin(y). from trig indentities: sin(y) = sqrt(1-cos^2(y)). Substituting gives dy/...
DP
5271 Views

You are given a polynomial f, where f(x)=x^4 - 14x^3 + 74 x^2 -184x + 208, you are told that f(5+i)=0. Express f as the product of two quadratic polynomials and state all roots of f.

Since x=5+i is a solution to f(x)=0 we then know that x=5-i must also be a solution to f(x)=0, by the complex conjugate root theorem.Now we can break f down into the product of a polynomial and these two kno...
PL
2612 Views

How do I find and plot the roots of a polynomial with complex roots on an Argand diagram? e.g. f(z) =z^3 -3z^2 + z + 5 where one of the roots is known to be 2+i

For a polynomial with real coefficients, use that roots come in complex conjugate pairs. Therefore, another root is 2-i (and we know for this example that the final root must be real). Write the factorised f...
ES
6086 Views

Using a Suitable substitution or otherwise, find the differential of y= arctan(sinxcosx), in terms of y and x.

First of all, replace sinxcosx with 1/2 sin2x. Then you should let U=1/2 Sin2x and replace that in the formula. If y=arctan(U), then U=tany. work out dU/dy which is Sec 2 y. Using the trigonometric identity ...
JP
3012 Views

Prove by induction that, for all integers n >=1 , ∑(from r=1 to n) r(2r−1)(3r−1)=(n/6)(n+1)(9n^2 -n−2). Assume that 9(k+1)^2 -(k+1)-2=9k^2 +17k+6

First, recall how to construct a proof by simple induction in this manner: (1) Assume statement true for n=k, (2) Prove true for n=k+1, (3) Show true for n=1.(1) => ∑(from r=1 to k) r(2r−1)(3r−1)= (k/6)(k...
TK
8044 Views