If we let u = 2x3 - 1 then y = u4.

So now

• y is written as a power of u; and
• u is a function of x [ u = f(x) ].

To find the derivative of such an expression, we can use our new rule:

`d/(dx)u^n=n u^(n-1)(du)/(dx`

where u = 2x3 − 1 and n = 4.

So

`(dy)/(dx)=n u^(n-1)(du)/(dx)`

`=[4(2x^3-1)^3][6x^2]`

`=24x^2(2x^3-1)^3`

We could, of course, use the chain rule, as before:

`(dy)/(dx)=(dy)/(du)(du)/(dx`
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.intmath.com/differentiation/ans-7.php?a=2", "fetch_time": 1516594198000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00411.warc.gz", "warc_record_offset": 917531368, "warc_record_length": 2118, "token_count": 179, "char_count": 413, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2018-05", "snapshot_type": "latest", "language": "en", "language_score": 0.7651284337043762, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"}