Help with finding a few derivatives?

Question

1.) x^8/cos(x)
2.) xe^2x
3.) 5t+6e^1.6t
4.) 9x^4/e^x
5.) ln(x^8+x)

Answer 1

(1)

x^8/cos(x) 

It can be written as x^8 *sec x

Apply the product rule of derivatives .

d (u v ) =  u v' + v u' 

Apply derivative with respect to x to x^8 *sec x .

d/dx (x^8 *sec x) = x^8 [sec x * tan x] + sec x [8*x^7]

                          = x^8 *sec x * tan x + sec x *8*x^7

                         = x^7*sec x [x tan x + 8 ]

                         = x^7 sec x [x tan x + 8 ]

So the derivative of x^8/cos(x) is x^7 sec x [x tan x + 8 ] .

Answer 2

(2)

xe^2x

Apply the product rule of derivatives .

d (u v ) =  u v' + v u' 

Apply derivative with respect to x to xe^2x .

d/dx (xe^2x) = x * e^2x  (2) + e^2x

                   =  x e^2x  (2) + e^2x

                    = e^2x [2x+ 1]

So the derivative of xe^2x  is  e^2x [2x+ 1] .

Answer 3

(3)

5t+6e^1.6t

Apply derivative with respect to t to 5t+6e^1.6t.

d/dt (5t+6e^1.6t) = 5 + 6 e^1.6t (1.6) 

                           = 5 + 9.6 e^1.6t                          [ since the derivative of t^n = n t^n-1  and e^(at) = a  e^(at) ]

So the derivative of  5t+6e^1.6t is 5 + 9.6 e^1.6t .

Answer 4

(4)

9x^4/e^x

It can be written as 9x^4e^(-x) .

Apply the product rule of derivatives .

d (u v ) =  u v' + v u' 

Apply derivative with respect to x to 9x^4e^(-x) .

d/dx (9x^4e^-x) = 9x^4 [e^-x (-1)] + e^-x[9*4x³]                       [ since the derivative of x^n = n x^n-1 ]

                          = -9x^4 e^-x  +36 x³ e^-x

                          =  9x³ e^-x [ -x + 4] 

So the deriavtive of 9x^4/e^x is 9x³ e^-x [ 4 -x ] .

Answer 5

(5)

ln(x^8+x)

Now derivative the function implicitly .

Derivative the function with respect to x .

d/dx (ln(x^8+x)) = 1/(x^8+x) * (x^8+x)'

                       = 1/(x^8+x) * (8x^7+1)

                      = (8x^7+1) / (x^8+x) 

So the derivative of the ln(x^8+x) is  (8x^7+1) / (x^8+x) .