Question

I'm looking for f(prime)(x)

http://webwork.math.ttu.edu/wwtmp/equations/96/9e19c155a90c794bf05182053f655d1.png

Answer 1

derivative?

Answer 2

product rule\[(uv)'=u'v+v'u\]

Answer 3

(2x^3-3)cos(x)+3x(2x+1)sin(x)

Answer 4

how were you able to take a image of your problem from webwork

Answer 5

i just clicked on it and dragged. which highlighted it then i copy and pasted image

Answer 6

so is that what the answer is jinnie?

Answer 7

yep, Junnie answer is correct!

Answer 8

can someone show me how that was done? I'm a little lost with derivative...

Answer 9

i got something way different

Answer 10

lets look only at
\[2x^3\sin(x)\] we use
\[(fg)'=f'g+g'f\] with
\[f(x)=2x^3, f'(x)=6x, g(x)=\sin(x), g'(x)=\cos(x)\] then plug directly in to the product rule

Answer 11

get
\[6x\sin(x)+2x^3\cos(x)\] by direct substitution. now we can do the second part in a similar fashion

Answer 12

The derivative of a sum of expressions is equal to the sum of the derivative of each expression. In other words, the property states that (f+g)'=f'+g'.

-3xcos(x)


Find the derivative of each expression, beginning with




Use the product rule to find the derivative of The product rule states that (gf)'=g'f+gf'.




The derivative of sin(x) is

sin(x)=cos(x)


Substitute the derivative back into the product rule formula.




The derivative of is




Substitute the derivative back into the product rule formula.




Simplify the derivative.




Next, find the derivative of -3xcos(x).

-3xcos(x)


Use the product rule to find the derivative of -3xcos(x). The product rule states that (hj)'=h'j+hj'.

cos(x)]


The derivative of -3x is

-3x=-3


Substitute the derivative back into the product rule formula.

cos(x)]


The derivative of cos(x) is

cos(x)=-sin(x)


Substitute the derivative back into the product rule formula.

-3xcos(x)=(-3)(cos(x))+(-3x)(-sin(x))


Simplify the derivative.

-3xcos(x)=-3cos(x)+3xsin(x)


The derivative of is




Since 2cos(x) and -3cos(x) are like terms, add -3cos(x) to 2cos(x) to get -cos(x).

Answer 13

We have
\[f(x)=2x^3\sin x-3x \cos x\]
Let's use the product rule here
\[f'(x)=\frac{d}{dx}(2x^3\sin x-3x \cos x)\]
we get now
\[f'(x)=\sin x\frac{d}{dx}(2x^3)+2x^3\frac{d}{dx}(\sin x)-(\cos x\frac{d}{dx}(3x )+3x\frac{d}{dx}(\cos x))\]
we get
\[f'(x)=\sin x \times (6x^2)+2x^3 \times \cos x-(3 cos x-3x \times \sin x)\]
we get
\[f'(x)=6x^2 \sin x+ 2x^3 \cos x-3 \cos x+3x \sin x\]

Answer 14

Do you have any question?

Answer 15

i got something way different