- #1
- 1,753
- 143
It's been too long since I've had an algebra class, so I start getting into trouble as these calculus questions rely more and more on algebra.
[tex]
\begin{array}{l}
y = (4x - 5)^4 \,(3x + 1)^5 \\
{\rm{Find the first derivative}}{\rm{. Simplify if possible (i}}{\rm{.e}}{\rm{. factor)}}{\rm{. Use the product rule}}{\rm{.}} \\
\left( {(4x - 5)^4 } \right)^\prime \left( {(3x + 1)^5 } \right) + \left( {(4x - 5)^4 } \right)\left( {(3x + 1)^5 } \right)^\prime \\
\\
\left( {4(4x - 5)^3 (4x - 5)'} \right)\left( {(3x + 1)^5 } \right) + \left( {(4x - 5)^4 } \right)\left( {5(3x + 1)^4 (3x + 1)'} \right) \\
\\
4(4x - 5)^3 (4)(3x + 1)^5 + (4x - 5)^4 \,\,5(3x + 1)^4 (3) \\
\\
16(4x - 5)^3 (3x + 1)^5 + 15(4x - 5)^4 (3x + 1)^4 \\
\end{array}
[/tex]
I only got 1 point out of 2 on this question. But when I graph it to check the answer, my formula seems to give me the correct slope at all points on the original function. Was the teacher expecting me to simplify this answer?
[tex]
\begin{array}{l}
y = (4x - 5)^4 \,(3x + 1)^5 \\
{\rm{Find the first derivative}}{\rm{. Simplify if possible (i}}{\rm{.e}}{\rm{. factor)}}{\rm{. Use the product rule}}{\rm{.}} \\
\left( {(4x - 5)^4 } \right)^\prime \left( {(3x + 1)^5 } \right) + \left( {(4x - 5)^4 } \right)\left( {(3x + 1)^5 } \right)^\prime \\
\\
\left( {4(4x - 5)^3 (4x - 5)'} \right)\left( {(3x + 1)^5 } \right) + \left( {(4x - 5)^4 } \right)\left( {5(3x + 1)^4 (3x + 1)'} \right) \\
\\
4(4x - 5)^3 (4)(3x + 1)^5 + (4x - 5)^4 \,\,5(3x + 1)^4 (3) \\
\\
16(4x - 5)^3 (3x + 1)^5 + 15(4x - 5)^4 (3x + 1)^4 \\
\end{array}
[/tex]
I only got 1 point out of 2 on this question. But when I graph it to check the answer, my formula seems to give me the correct slope at all points on the original function. Was the teacher expecting me to simplify this answer?