Question

if f(x) + x²f(x)³ = 10 and f(1) = 2, find f(1).
f(1) =

1. -/1 points

use implicit differentiation to find an equation of the tangent line to the curve at the given point.
y sin(12x) = x cos(2y), (π/2, π/4)
y =

Explanation:

Step1: Differentiate both sides

Differentiate $f(x)+x^{2}[f(x)]^{3}=10$ with respect to $x$ using sum - rule, product - rule and chain - rule.
The derivative of $f(x)$ is $f^{\prime}(x)$. For the second term $x^{2}[f(x)]^{3}$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v=[f(x)]^{3}$. The derivative of $u=x^{2}$ is $u^\prime = 2x$, and the derivative of $v=[f(x)]^{3}$ using the chain - rule is $v^\prime=3[f(x)]^{2}f^{\prime}(x)$. So the derivative of $x^{2}[f(x)]^{3}$ is $2x[f(x)]^{3}+3x^{2}[f(x)]^{2}f^{\prime}(x)$. The derivative of the right - hand side (a constant 10) is 0. So we have $f^{\prime}(x)+2x[f(x)]^{3}+3x^{2}[f(x)]^{2}f^{\prime}(x)=0$.

Step2: Substitute $x = 1$ and $f(1)=2$

Substitute $x = 1$ and $f(1)=2$ into $f^{\prime}(x)+2x[f(x)]^{3}+3x^{2}[f(x)]^{2}f^{\prime}(x)=0$.
We get $f^{\prime}(1)+2\times1\times[2]^{3}+3\times1^{2}\times[2]^{2}f^{\prime}(1)=0$.
Which simplifies to $f^{\prime}(1)+16 + 12f^{\prime}(1)=0$.

Step3: Solve for $f^{\prime}(1)$

Combine like terms: $(1 + 12)f^{\prime}(1)=-16$.
$13f^{\prime}(1)=-16$.
So $f^{\prime}(1)=-\frac{16}{13}$.

Answer:

$-\frac{16}{13}$