Find dy.
-$$y = \ln \left( 1 + x ^ { 4 } \right)$$
A) $\left( \frac { 4 x ^ { 3 } } { x ^ { 4 } + 1 } \right) d x$
B) $\left( \frac { x ^ { 3 } } { x ^ { 4 } + 1 } \right) d x$
C) $\left( \frac { 4 x ^ { 3 } } { x ^ { 4 } - 1 } \right) d x$
D) $\left( \frac { x ^ { 4 } } { x ^ { 4 } + 1 } \right) d x$

Question

Quizplus · Accepted Answer

To find $dy$, we differentiate $y = \ln(1 + x^4)$ with respect to $x$. Using the chain rule, the derivative is $\frac{1}{1 + x^4} \cdot 4x^3$, which simplifies to $\frac{4x^3}{1 + x^4}$. Therefore, $dy = \left( \frac{4x^3}{x^4 + 1} \right) dx$.

Find Dy - y=ln⁡(1+x4)y = \ln \left( 1 + x ^ { 4 } \right)y=ln(1+x4)

Find Dy
- $y = \ln \left( 1 + x ^ { 4 } \right)$