Derivatives of y =(x²+1) (x³+3).
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From the product rule with u=x
2
+1 and v=x
3
, we find
dx
d
[(x
2
+1)(x
3
+3)]=(x
2
+1)
dx
d
(x
3
+3)+(x
3
+3)
dx
d
(x
2
+1)
(x
2
+1)(3x
2
)+(x
3
+3)(2x)
=3x
4
+3x
2
+2x
4
+6x
=5x
4
+3x
2
+6x.
The above sum can be done as well (perhaps better multiplying out the original expression for y and differential the resulting polynomial. We now check;
y=(x
2
+1)(x
3
+3)=x
5
+x
3
+3x
2
+3
dx
dy
=5x
4
+3x
2
+6x
This is in agreement with our first calculation.
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