calc help g(-2)=-1, g(3)=1, g(1)=6, g'(-2)=2, g'(3)=3, g'(1)=5?

Question

find
a. if H(x)=ln(g(x)+e^(2x)) then H'(3)=
b. if K(x) = arcsin(g(x)+1) then k'(-2)=
c. if M(x)=g(x)arctan x then m'(1)=

pleas explain how to do this!!

Answer 1

a. H'(x) = [1/(g(x)+e^(2x))][g'(x) + 2e^(2x)]

H'(3) = [1/(1+e^6)][3 + 2e^6]



b. K'(x) = g'(x)/sqrt[1 - (g(x)+1)^2]

K'(-2) = 2/sqrt[1 - (-1+1)^2] = 2



c. M'(x) = g'(x) arctan x + g(x)/(1+x^2)

M'(1) = 5(pi/4) + 6/(1+1) = (5/4)pi + 3

Answer 2

h(x) = ln(g(x) + e^(2x))

h'(x) = (g'(x) + 2 * e^(2x)) / (g(x) + e^(2x))

h'(3) = (g'(3) + 2 * e^(2 * 3)) / (g(3) + e^(2 * 3))

h'(3) = (3 + 2 * e^6) / (1 + e^6)



k(x) = arcsin(g(x) + 1)

sin(k(x)) = g(x) + 1

cos(k(x)) * k'(x) = g'(x)

k'(x) = g'(x) / cos(k(x))

k'(x) = g'(x) / sqrt(1 - sin(k(x))^2)

k'(x) = g'(x) / sqrt(1 - sin(arcsin(g(x) + 1))^2)

k'(x) = g'(x) / sqrt(1 - (g(x) + 1)^2)

k'(-2) = g'(-2) / sqrt(1 - (g(-2) + 1)^2)

k'(-2) = 2 / sqrt(1 - (-1 + 1)^2)

k'(-2) = 2 / sqrt(1 - 0)

k'(-2) = 2 / 1

k'(-2) = 2





m(x) = g(x) * arctan(x)

m'(x) = g(x) * 1 / (1 + x^2) + arctan(x) * g'(x)

m'(1) = g(1) / (1 + 1^2) + arctan(1) * g'(1)

m'(1) = 6 / (1 + 1) + (pi/4) * 5

m'(1) = 3 + 5pi/4

Answer 3

a) using d/du(lnu) = u'/u

H(x)=ln(g(x)+e^(2x))

H'(x)=(g'(x)+2e^(2x))/(g(x)+e^(2x))

H'(3)=(g'(3)+2e^(2*3))/(g(3)+e^(2*3))

=(3+2e^6)/(1+e^6)



b) using d/du(arcsinu)=u'/sqrt(1-u^2)

K(x) = arcsin(g(x)+1)

k'(x)=g'(x)/sqrt(1-(g(x)+1)^2)

k'(-2)=g'(-2)/sqrt(1-(g(-2)+1)^2)

=2/sqrt(-8) = does not exist



c) M(x)=g(x)arctan x

M'(x) = g(x)/(1+x^2) + g'(x)arctanx

m'(1) = g(1)/(1+1^2) + g'(1)arctan1

=3+5*pi/4