Let $$ y=12x^4\sqrt[3]{x^2}-2e^x+9 $$ How can we find $y^\prime$?
Asked
Active
Viewed 91 times
1
-
Are you asked to find the derivative? – Paul Sundheim Aug 25 '14 at 01:26
-
Yes, prime derivative of 'y'. – Peep Aug 25 '14 at 01:28
-
Using the table derivatives. – Peep Aug 25 '14 at 01:29
-
2To begin with try combining the powers $x^4(x^2)^{\frac{1}{3}}$ – Paul Sundheim Aug 25 '14 at 01:34
-
You will almost surely need to use more than one of the formulas from your table of derivatives. – David K Aug 25 '14 at 02:09
1 Answers
3
$$\require{cancel} y=12x^4\underbrace{\sqrt[3]{x^2}}_{x^\frac23}-2e^x+9=12x^\frac{14}3-2e^x+9\\\frac{dy}{dx}=\cancelto{4}{12}\left(\frac{14}{\cancel{3}}\right)(x^\frac{11}3)-2e^x+0\\=56x^\frac{11}3-2e^x$$
Shahar
- 3,302