Question

Seeking for the right answer..help me please.. The question is about differentiate with respect to x. Given, y=e with power 3x cosh x square

Answer 1

do you mean \[y = e^{3x}\cosh x^2\]or\[y = e^{3x}\cosh^2 x\]?

Answer 2

yes

Answer 3

is the square on the coshine or on the x ?

Answer 4

on the x

Answer 5

ok so
\[y = e^{3x}\cosh x^2\]
and you have to take the derivative with respect to \(x\)

Answer 6

Do you remember the product rule of differentiation?

Answer 7

is it differentiate both? so, dy/dx =e^3x and d/dx cosh x^2

Answer 8

The product rule states that
if a function \(y(x)\) is composed of a product of functions \(f(x),\, g(x)\)\[y(x)=f\cdot g\]
The derivative will be equal to the sum:
\[y'= f'g+fg'\]

Answer 9

so let \(f(x) = e^{3x}\)
and find the derivative \(f'= \quad. . .\)

and let \(g(x)=\cosh(x^2)\)
and find its derivative also \(g' = \quad...\)

Answer 10

if we differentiate cosh x^2, is it get sinh x^2?

Answer 11

and differentiate e^3x will remain same, e^3x?

Answer 12

dont forget the chain rule
d/dx ( cosh (x^2) ) = sinh (x^2) * d/dx ( x^2 )

and also
d/dx ( e^3x ) = e^3x * d/dx ( 3x )

Answer 13

ohh i see..

Answer 14

\[y(x)=(f\circ g)(x)=f(g(x))\]
when taking the derivative of a function of a function, you have to multiply the derivative of the outside function, by the derivative of the inside function

\[y'=(f'\circ g)\cdot g'\]

Answer 15

for example
\[y = e^{x^2}\\
y'=2x\cdot e^{x^2}\]

Answer 16

i see.. if differentiate e^x will remain same .. is it right?

Answer 17

yes
d/dx ( e^x ) = e^x * d/dx ( x )
= e^x * 1
= e^x

Answer 18

i see.. so the answer for this question is 3e^3x sinh x^2 x^2

Answer 19

hehe is it right? or can simply?

Answer 20

ehh sorry, the last is 2x, not x^2

Answer 21

From the product formula\[y=f\cdot g\implies y'= f'g+fg'\]
we let \(f=e^{3x}\) and \(g=\cosh(x^2)\)

what did you get for \(f'\) and \(g'\) ?

Answer 22

f' = 3e^3x and g' = 2x sinh x^2

Answer 23

that's right!
now the derivative of the function \(y\) is \[y'= f'g+fg'\]

Answer 24

yess.. need to plus both?

Answer 25

\[y'=3e^{3x}\cdot\cos(x^2)+e^{3x}\cdot2x\sinh(x^2)\]

Answer 26

you can simplify this if you factor it a little

Answer 27

how..? i can't see :D

Answer 28

there is a common factor to both terms in the sum

Answer 29

e^3x (3cos(x^2) + 2xsinh(x^2) ?

Answer 30

x^2 can taken out?

Answer 31

That's it, your done

Answer 32

the x^2's are trapped the hyperbolic functions

Answer 33

thats all? hee thanks a lot for help ^^

Answer 34

i see