Solve d/dx(x^2+x-5)^3

$$\frac { d } { d x } ( x ^ { 2 } + x - 5 ) ^ { 3 }$$

Evaluate

$$3\left(2x+1\right)\left(x^{2}+x-5\right)^{2}$$

Steps Using Chain Rule

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$3\left(x^{2}+x^{1}-5\right)^{3-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+x^{1}-5)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$3\left(x^{2}+x^{1}-5\right)^{2}\left(2x^{2-1}+x^{1-1}\right)$$

Simplify.

$$\left(x^{2}+x^{1}-5\right)^{2}\left(6x^{1}+3x^{0}\right)$$

For any term t, t^{1}=t.

$$\left(x^{2}+x-5\right)^{2}\left(6x+3x^{0}\right)$$

For any term t except 0, t^{0}=1.

$$\left(x^{2}+x-5\right)^{2}\left(6x+3\times 1\right)$$

For any term t, t\times 1=t and 1t=t.

$$\left(x^{2}+x-5\right)^{2}\left(6x+3\right)$$

Differentiate w.r.t. x

$$30x^{4}+60x^{3}-144x^{2}-174x+120$$