What is the derivative of y=(k(x)^2+x^2)^4(x+k(x)) ?

The derivative of y=(k(x)^2+x^2)^4(x+k(x)) is 4(kx^2+x^2)^3(2kx+2x)(x+kx)+(1+k)(kx^2+x^2)^4

What is the first derivative of y=(k(x)^2+x^2)^4(x+k(x)) ?

The first derivative of y=(k(x)^2+x^2)^4(x+k(x)) is 4(kx^2+x^2)^3(2kx+2x)(x+kx)+(1+k)(kx^2+x^2)^4

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derivative of

y = (k (x)^{2} + x^{2})^{4} (x + k (x))

4 (k x^{2} + x^{2})^{3} (2 k x + 2 x) (x + k x) + (1 + k) (k x^{2} + x^{2})^{4}

Solution steps

\frac{d}{d x} ((k x^{2} + x^{2})^{4} (x + k x))

Apply the Product Rule:

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

f = (k x^{2} + x^{2})^{4}, g = x + k x

= \frac{d}{d x} ((k x^{2} + x^{2})^{4}) (x + k x) + \frac{d}{d x} (x + k x) (k x^{2} + x^{2})^{4}

\frac{d}{d x} ((k x^{2} + x^{2})^{4}) = 4 (k x^{2} + x^{2})^{3} (2 k x + 2 x)

\frac{d}{d x} (x + k x) = 1 + k

= 4 (k x^{2} + x^{2})^{3} (2 k x + 2 x) (x + k x) + (1 + k) (k x^{2} + x^{2})^{4}

What is the derivative of y=(k(x)^2+x^2)^4(x+k(x)) ?
The derivative of y=(k(x)^2+x^2)^4(x+k(x)) is 4(kx^2+x^2)^3(2kx+2x)(x+kx)+(1+k)(kx^2+x^2)^4
What is the first derivative of y=(k(x)^2+x^2)^4(x+k(x)) ?
The first derivative of y=(k(x)^2+x^2)^4(x+k(x)) is 4(kx^2+x^2)^3(2kx+2x)(x+kx)+(1+k)(kx^2+x^2)^4