e^(6) is merely a style (approx 403, do it on your calculator). so which you would be able to think of of the question as being some thing like: e^(7x) - 403 The spinoff of a style is 0, subsequently the spinoff of the question is: 7e^(7x) i'm a severe college maths instructor and the respond is 7e^(7x) you're top: the rule of derivitives is that der.e^x=e^x. even with the undeniable fact that, it somewhat is barely for e^x. listed right here are some examples of derivatives e^2x ? 2e^2x e^x² ? 2xe^x² e^(x² - 3x) ? (2x -3)e^(x² - 3x) so which you're able to desire to multiply e by employing the differential of the ability.
Comments
First use the product rule and you get
x'cos^2x + x(cos^2x)'
x' = 1, and apply the chain rule on the second term:
cos^2x - 2xsinx
e^(6) is merely a style (approx 403, do it on your calculator). so which you would be able to think of of the question as being some thing like: e^(7x) - 403 The spinoff of a style is 0, subsequently the spinoff of the question is: 7e^(7x) i'm a severe college maths instructor and the respond is 7e^(7x) you're top: the rule of derivitives is that der.e^x=e^x. even with the undeniable fact that, it somewhat is barely for e^x. listed right here are some examples of derivatives e^2x ? 2e^2x e^x² ? 2xe^x² e^(x² - 3x) ? (2x -3)e^(x² - 3x) so which you're able to desire to multiply e by employing the differential of the ability.
xcos^2(x)
cos^2 (x) + x * 2 cos x sin x
cos^2 (x) + 2x sinx cosx
cos^2 (x) + x sin(2x)
solved!!!!
using d/dx of (uv) = vdu + udv and d/dx cos = -sin
d/dx xcos(x)^2 = cos(x)^2 - 2xcos(x)sin(x)
x * cos(x)^2
f'(x) = x'cos(x)^2 + x(cos(x)^2)'
f'(x) = cos(x)^2 + x(-2cos(x)sin(x))
f'(x) = cos(x)^2 - (2x)cos(x)sin(x)
you can also shorten this to
f'(x) = cos(x)^2 - xsin(2x)