Question

Suppose the derivative of f exists, and assume that f(1)=3, and f′(1)=5. Let g(x)=x^2 f(x), and h(x)=[f(x)]/[x−0].

a) g′(1)=..........................
Find the equation of the tangent line to g(x) at x=1.
y=..........................

b) h′(1)=..........................
Find the equation of the tangent line to h(x) at x=1.
y=...........................


Hint: Find g′(x) and h′(x) in terms of f(x) and f′(x). You just need to use the basic product and quotient rules. The equation of a tangent line to a function k(x) at x=x0 is: y=k′(x0)(x−x0)+k(x0).

Answer 1

for g'(x) you'll need to product rule... do you know it..?

Answer 2

product rule of x^2 = 2x?

Answer 3

so do you know the derivative of g'(x) ... and thats only part of the product rule.

Answer 4

can you tell me the equation i am supposed to find derivative of?

Answer 5

but, use the f(x) as your (x,y) (1,3)

Hope this helps.

Answer 6

ok... for g(x) let u = x^2 du/dx = 2x
v = f(x) du/dx = f'(x)

so using the product rule

\[g'(x) = u \times \frac{dv}{dx} + v \times \frac{du}{dx}\]

so you will have

\[g'(x) = x^2 \times f'(x) + 2x \times f(x)\]

you need to find \[g'(1) = (1)^2 \times f'(1) + 2(1) \times f(1)\]

f(1) and f'(1) are given above... substitute and evaluate which will give the gradient at x = 1

to find your point evaluate

g(1) = (1)^2 x f(1)

so you point will be (1 , g(1))

then find the equation.

Answer 7

so g'(1) = 1 x 5 + 2 x 3=11

Answer 8

that correct... so thats the slope at x = 1

you now need to find g(1) ...

g(1) = (1)^2 * f(1)

Answer 9

g(1) = 3

Answer 10

ok... so you need to find the equation of a straight line( the tangent) with m = 11 and the point (1, 3)

can you do that..?

Answer 11

y-3=11(x-1)?

Answer 12

33x-33

Answer 13

isn't it

y - 3 = 11(x - 3)

so y - 3 = 11x - 33

or

y = 11x - 30

Answer 14

ready for h(x)?

Answer 15

yup. all processed

Answer 16

so should it be:
f'(x)(x-0)-f(x)(x-0)/(x-0)^2?

Answer 17

ok... so for the quotient rule

ok... almost correct... you just need the derivative of x - 0

Answer 18

yes yes of course

Answer 19

so you will have

\[h'(1) = \frac{ f'(1)\times (1 - 0) - f(1) \times 1}{(1 - 0)^2}\]

Answer 20

so.....-10/1?

Answer 21

=-10

Answer 22

ok... so thats the gradient...

you need to find h(1) as the point for the line is (1, h(1))

Answer 23

just a wild guess:
will it be 3/(1-0)?

Answer 24

=3?

Answer 25

ok.. so find the equation of the tangent where m = -10 and the point is (1, 3)

can you do that..?

Answer 26

y-3= -10(x-3)?

Answer 27

=-30x=90?

Answer 28

ok... thats great

Answer 29

so

y - 3 = -10(x - 1)

y - 3 = -10x + 10

Answer 30

then y = -10x + 13

should be the equation of the tangent

Answer 31

the computer shows that the only right answer i got out of the 4 is the first one.g'(1)=11

Answer 32

y=11x-30 (incorrect)
h'(1)= -10 (incorrect)
y=-10x+13(incorrect)

Answer 33

ok... the tangent for the 1st line is

y - 3 = 11(x - 1)

so

y -3= 11x - 11

y = 11x - 8

try than for the tangent to g(x)

Answer 34

you got it right

Answer 35

and for h(x) you should have

\[h'(1) = \frac{( 1- 0) time 5 - (3 \times 1)}{1} = 2\]


so the value of h'{1} = 2

and

y - 3 = 2( x - 1)

y - 3 = 2x - 2

y = 2x + 1 is the tangent

Answer 36

YES! YES! YES!

Answer 37

thank you very much campbell

Answer 38

ok... just arithmetic errors...

hope it all helped.

Answer 39

yes, very much

Answer 40

you want to work on more questions maybe?

Answer 41

nope... I think you need to go and practice the product and quotient rules...

Answer 42

ok great. thanks again and see you around