What is the (f(x)(x^2))' ?

The (f(x)(x^2))' is f'(x)x^2+2xf(x)

What is the first (f(x)(x^2))' ?

The first (f(x)(x^2))' is f'(x)x^2+2xf(x)

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(f (x) (x^{2}))^{'}

f' (x) x^{2} + 2 x f (x)

Solution steps

(f (x) (x^{2}))^{'}

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = f (x), g = x^{2}

= f^{^{'}} (x) x^{2} + (x^{2})^{^{'}} f (x)

(x^{2})^{'} = 2 x

= f^{^{'}} (x) x^{2} + 2 x f (x)

f (x) = \frac{sin ( x )}{cos ( ln ( - x ) )}

x y - 4 y^{2} = 8

\int 5 x^{3} d x

\int \frac{( 2 t ^{2} + t t - 1 )}{t ^{2}} d t

\int \frac{1}{sin ( y ) + 1} d y