Mixing
Calc2 Finding $f'(2)$ from tangent line
$begingroup$
So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$
Any ideas on how to go about this? Thank you!
derivatives tangent-line
edited Jan 25 at 3:16
Christopher Marley
1,129115
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
$endgroup$
add a comment |
$begingroup$
So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$
Any ideas on how to go about this? Thank you!
derivatives tangent-line
edited Jan 25 at 3:16
Christopher Marley
1,129115
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
$endgroup$
1
$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09
add a comment |
$begingroup$
So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$
Any ideas on how to go about this? Thank you!
derivatives tangent-line
edited Jan 25 at 3:16
Christopher Marley
1,129115
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
$endgroup$
So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$
Any ideas on how to go about this? Thank you!
derivatives tangent-line
derivatives tangent-line
edited Jan 25 at 3:16
Christopher Marley
1,129115
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
edited Jan 25 at 3:16
Christopher Marley
1,129115
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
edited Jan 25 at 3:16
Christopher Marley
1,129115
edited Jan 25 at 3:16
Christopher Marley
1,129115
edited Jan 25 at 3:16
Christopher Marley
1,129115
1,129115
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
asked Jan 25 at 3:05
Jelly BiscuitJelly Biscuit
204
204
1
$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09
add a comment |
1
$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09
1
1
$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09
$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.
Now, $$f(x) = -3[h(x)]^2+2x+2$$
By chain rule for h(x)
$$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
x=-2
$$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
Simplify
$$f'(-2) = -6cdot 4cdot 3+2 = -70$$
edited Jan 25 at 4:13
El borito
666216
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
$endgroup$
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
add a comment |
$begingroup$
We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.
Now, $$f(x) = -3[h(x)]^2+2x+2$$
By chain rule for h(x)
$$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
x=-2
$$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
Simplify
$$f'(-2) = -6cdot 4cdot 3+2 = -70$$
edited Jan 25 at 4:13
El borito
666216
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
$endgroup$
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
add a comment |
$begingroup$
Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.
Now, $$f(x) = -3[h(x)]^2+2x+2$$
By chain rule for h(x)
$$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
x=-2
$$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
Simplify
$$f'(-2) = -6cdot 4cdot 3+2 = -70$$
edited Jan 25 at 4:13
El borito
666216
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
$endgroup$
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
add a comment |
$begingroup$
Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.
Now, $$f(x) = -3[h(x)]^2+2x+2$$
By chain rule for h(x)
$$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
x=-2
$$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
Simplify
$$f'(-2) = -6cdot 4cdot 3+2 = -70$$
edited Jan 25 at 4:13
El borito
666216
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
$endgroup$
Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.
Now, $$f(x) = -3[h(x)]^2+2x+2$$
By chain rule for h(x)
$$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
x=-2
$$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
Simplify
$$f'(-2) = -6cdot 4cdot 3+2 = -70$$
edited Jan 25 at 4:13
El borito
666216
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
edited Jan 25 at 4:13
El borito
666216
edited Jan 25 at 4:13
El borito
666216
edited Jan 25 at 4:13
El borito
666216
666216
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
answered Jan 25 at 3:13
Christopher MarleyChristopher Marley
1,129115
1,129115
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
add a comment |
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
$begingroup$
thank you so much!
$endgroup$
– Jelly Biscuit
Jan 25 at 5:46
add a comment |
$begingroup$
We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
add a comment |
$begingroup$
We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
add a comment |
$begingroup$
We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
edited Jan 25 at 3:16
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
answered Jan 25 at 3:10
Jimmy SabaterJimmy Sabater
3,023325
3,023325
add a comment |
add a comment |
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Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09