Verified answer

f(x) = (3x^2 + 5)^7(−4x^2 + 7)^11

Differentiation by product rule (2 functions of x):

u= (3x^2 + 5)^7

du/dx = 7*6x(3x^2 + 5)^6

v=(-4x^2 + 7)^11

dv/dx = 11*-8x(-4x^2+7)^10

f'(x) = (42x(3x^2+5)^6)(-4x^2+7)^11) - (88x(-4x^2+7)^10)((3x^2+5)^7)

2. f(x) = (x^2+ 2x + 7)^4

using chain rule:

f'(x) = 4(2x+2)(x^2 + 2x +7)^3

f'(5) = 4(10+2)(5^2 + 10 +7)^3 = 4*12(25+17)^3 = 4*12*42^3

= 3 556 224.

f(x) or g(x) or y(x) or z(x) are referred to as applications. this implies, once you plug in a quantity or fee in for x, the function will spit out yet another fee. think of of it as a device, this device is named f(x). For in spite of fee "x" you place into the device, f(x) will spit out something bearing directly to that x. we could circulate by way of an occasion. we've a f(x) device which will continually double your enter. nicely how do you describe this device in words? f(x)=2x. which ability, for in spite of x fee you put in, f(x) will continually be 2 cases that fee. in case you put in 2 for x, f(2)=2(2) so f(2) equals 4. So on your situation, you have a f(x)=3x+4 and a g(x)=2x. for this reason, you have if truth be told 2 machines, the f(x) and the g(x). we could use the device analogy lower back and address the 1st section. 2f(5)=what? nicely, what's this truly asking? First you are able to desire to settle on which device this situation is speaking approximately. 2f(5) needs you to %. out the f(x) device particularly than the g(x) device. 2f(5) states that the "x" fee for this device will might desire to be 5. for this reason, once you place a cost of 5 into the f(x) device, you get f(x)=3x+4, 3(5)+4, which equals 19. for this reason, f(5) equals 19. even inspite of the undeniable fact that, they desire 2 cases f(5) so 2 cases 19 is 38. with a bit of luck it truly is sensible. We do the comparable element for the 2nd section. f(2) refers back to the f(x) device. 3(2)+4 equals 10. g(2) refers back to the g(x) device. 2(2) equals 4. Then merely use user-friendly multiplication. 10 cases 4 equals 40. There you circulate, your solutions are 38 and 40.