# Differentiate y= x^3 * tanx

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y= x^3 * tanx

We note that we had a product of two functions.

Then we will use the product rule to solve:

Let y= u*v such that:

u= x^3 ==> u' = 3x^2

v= tanx ==> v' = sec^2 x

Then we know that:

y' = u'v + uv'

Let us substitue with u and v values:

==> y' = 3x^2 * tanx + x^3 * sec^2 x

Now we will factor x^2:

**==> y' = x^2(3tanx + xsec^2 x)**

To differentiate y = x^3 tanx.

The right side is a product of x^3 and tanx. So we use the product rule to differentiate:

(u(x)(v(x)}' = u'(x)v(x) +u(x)v'(x).

Here u(x) = x^3 , u'(x) = (x^3)' = 3x^2

v(x) = tanx . (v(x))' = (tanx)' = sec^2x.

Therefore y' = (x^3*tanx)'.

y'(x) = (x^3)'tanx +x^3(tanx)'.

y' = 3x^2tanx+x^3* sec^2x.

y' = x^2 {3tanx+xsec^2 x) .