How do you find the derivative of #x^3*arctan(7x)#?

Question

William Abdalla · Accepted Answer

Firstly let's differentiate this function using implicit and logarithmic differentiation:

#q=ln(arctan(7x))#

#e^q=arctan(7x)#

#tan(e^q)=7x#

#e^qsec^2(e^q)*(dq)/(dx)=7#

#arctan(7x)*(tan^2(e^q)+1)(dq)/(dx)=7#

#arctan(7x)*(49x^2+1)(dq)/(dx)=7#

#(dq)/(dx)=7/(arctan(7x)*(49x^2+1)#

Alright, knowing this we can now differentiate #x^3*arctan(7x)# using implicit differentiation and the result above...

#y=x^3*arctan(7x)#

#lny=ln(x^3*arctan(7x))#

#lny=ln(x^3)+ln(arctan(7x))#

#lny=3lnx+ln(arctan(7x))#

#1/y*(dy)/(dx)=3/x+7/(arctan(7x)*(49x^2+1)#

#(dy)/(dx)=y{3/x+(7)/(arctan(7x)*(49x^2+1)}}#

#(dy)/(dx)=x^3*arctan(7x){3/x+7/(arctan(7x)*(49x^2+1)}}#

#(dy)/(dx)=(3x^3*arctan(7x))/x+(7x^3*arctan(7x))/(arctan(7x)*(49x^2+1))#

#(dy)/(dx)=3x^2*arctan(7x)+(7x^3)/(49x^2+1)#

I can also give you an alternative way of finding this derivative, using the product rule...

#y=x^3*arctan(7x)=u*v#

#u=x^3#, therefore #(du)/(dx)=3x^2#

#v=arctan(7x)#

#tanv=7x#

#sec^2v*(dv)/(dx)=7#

#(tan^2v+1)*(dv)/(dx)=7#

#(49x^2+1)*(dv)/(dx)=7#

#(dv)/(dx)=7/(49x^2+1)#

This means that:

#(dy)/(dx)=x^3*7/(49x^2+1)+arctan(7x)*3x^2#

#(dy)/(dx)=(7x^3)/(49x^2+1)+3x^2*arctan(7x)#

Christopher Allers · Answer

undefined To find the derivative of $x^3 \cdot \arctan(7x)$, you can use the product rule of differentiation. The product rule states that if $u(x)$ and $v(x)$ are differentiable functions of $x$, then the derivative of their product is given by $(u \cdot v)' = u'v + uv'$.

Let $u(x) = x^3$ and $v(x) = \arctan(7x)$.

Now, $u'(x) = 3x^2$ (derivative of $x^3$) and $v'(x) = \frac{7}{1 + (7x)^2}$ (derivative of $\arctan(7x)$ using the chain rule).

Applying the product rule:

$$
\begin{align*}
(uv)' &= u'v + uv' \
&= (3x^2) \cdot \arctan(7x) + x^3 \cdot \frac{7}{1 + (7x)^2} \
&= 3x^2 \cdot \arctan(7x) + \frac{7x^3}{1 + 49x^2}.
\end{align*}
$$

So, the derivative of $x^3 \cdot \arctan(7x)$ is $3x^2 \cdot \arctan(7x) + \frac{7x^3}{1 + 49x^2}$.