In the equation below,
𝐶
is some constant.

13
𝑥
𝑒
𝑦
−
3
𝑦
=
𝐶
+
4
𝑥

Suppose the equation represents a curve passing through
(
2
,
−
1
)
. Evaluate
𝑦
′
at
(
2
,
−
1
)
.

Question

In the equation below, 
𝐶
 is some constant.

13
𝑥
𝑒
𝑦
−
3
𝑦
=
𝐶
+
4
𝑥

Suppose the equation represents a curve passing through 
(
2
,
−
1
)
. Evaluate 
𝑦
′
 at 
(
2
,
−
1
)
.

BlackTom AI · Accepted Answer

Differentiate implicitly with respect to x.

Relevant concepts: product rule (d(uv)/dx = u'v + uv'), chain rule for functions of y(x) (d/dx [y] = y'), and constants differentiate to zero.

Start from the interpreted equation: 13x e^y − 3y = C + 4x. Differentiate each term:
- d/dx[13x e^y] = 13 * d/dx[x e^y] = 13[(d/dx x) e^y + x (d/dx e^y)] = 13[e^y + x e^y y'] = 13 e^y + 13 x e^y y'.
- d/dx[−3y] = −3 y'.
- d/dx[C + 4x] = 0 + 4.

So the differentiated equation is:
13 e^y + 13 x e^y y' − 3 y' = 4.

Collect y' terms:
y'(13 x e^y − 3) + 13 e^y = 4.

Solve for y':
y'(13 x e^y − 3) = 4 − 13 e^y,
so y' = (4 − 13 e^y) / (13 x e^y − 3).

Evaluate at the point (x, y) = (2, −1):
Compute intermediate values:
e^y = e^{−1} = 1/e.
13 x e^y = 13·2·e^{−1} = 26 e^{−1}.

Plug in:
Numerator = 4 − 13 e^{−1}.
Denominator = 26 e^{−1} − 3.

Therefore y' at (2, −1) is (4 − 13 e^{−1}) / (26 e^{−1} − 3).

Why this is correct: the product rule produced the 13 e^y and 13 x e^y y' terms; the chain rule produced −3 y'; solving algebraically isolates y' and substitution gives the stated rational expression. There are no alternative options listed to compare; any incorrect alternative would have omitted product or chain rule contributions or evaluated exponentials incorrectly.

Final confirmation: y'(2, −1) = (4 − 13 e^{−1}) / (26 e^{−1} − 3), which matches the provided answer (written there as numerator 4 − 13 e^{−1} over denominator −3 + 26 e^{−1}).

In the equation below, 𝐶 is some constant. 13 𝑥 𝑒 𝑦 − 3 𝑦 = 𝐶 + 4 𝑥 Suppose the equation represents a curve passing through ( 2 , − 1 ) . Evaluate 𝑦 ′ at ( 2 , − 1 ) .单项选择题

类似问题

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