Solve f(t)=(9t+7)^2/3

$$f ( t ) = ( 9 t + 7 ) ^ { 2 / 3 }$$

Differentiate w.r.t. t

$$\frac{6}{\sqrt[3]{9t+7}}$$

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).

$$\frac{2}{3}\left(9t^{1}+7\right)^{\frac{2}{3}-1}\frac{\mathrm{d}}{\mathrm{d}t}(9t^{1}+7)$$

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}.

$$\frac{2}{3}\left(9t^{1}+7\right)^{-\frac{1}{3}}\times 9t^{1-1}$$

Simplify.

$$6t^{0}\left(9t^{1}+7\right)^{-\frac{1}{3}}$$

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

$$6t^{0}\left(9t+7\right)^{-\frac{1}{3}}$$

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

$$6\times 1\left(9t+7\right)^{-\frac{1}{3}}$$

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

$$6\left(9t+7\right)^{-\frac{1}{3}}$$

Evaluate

$$\left(9t+7\right)^{\frac{2}{3}}$$