Derivative problem

[notadummy]

fine the derivative of the function

g(t)=t^3(t^4+5)^7/2

the problem is "t" raised to the 3rd power times t raised to the 4th power plus five raised to the 7/2 power

 

[Gene]

t^3(t^4+5)^(7/2)
Use the product rule.
d(a*b) = a*db + b*da
t^3(d(t^4+5)^(7/2) + (t^4+5)^(7/2)d(t^3)

 

[soroban]

Hello, notadummy!

Exactly where is your difficulty?
. . The product rule?
. . The chain rule?
. . Basic algebra?

The most likely scenario: The book has a totally <u>simplified</u> answer
. . and you want to know how they did it.

Find the derivative: .$\displaystyle g(t)\:=\:t^3(t^4\,+\,5)^{\frac{7}{2}}$

Differentiate: .$\displaystyle g'(x)\:=\:t^3\cdot\frac{7}{2}(t^4\,+\,5)^{\frac{5}{2}}\cdot4t^3\:+\:3t^2\cdot(t^4\,+\,5)^{\frac{7}{2}}$

. . . . $\displaystyle =\;14t^6(t^4\,+\,5)^{\frac{5}{2}}\:+\:3t^2(t^4\,+\,5)^{\frac{7}{2}}$

Factor: .$\displaystyle t^2\cdot(t^4\,+\,5)^{\frac{5}{2}}\cdot[14t^4\,+\,3(t^4\,+\,5)]$

. . . . $\displaystyle =\;t^2\cdot(t^4\,+\,5)^{\frac{5}{2}}\cdot[14t^4\,+\,3t^4\,+\,15]$

. . . . $\displaystyle =\;t^2\cdot(t^4\,+\,5)^{\frac{5}{2}}\cdot(17t^4\,+\,15)$