Solve (2t+3)^1/2

$$( 2 t + 3 ) ^ { 1 / 2 }$$

Differentiate w.r.t. t

$$\frac{1}{\sqrt{2t+3}}$$

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).

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

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}.

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

Simplify.

$$t^{0}\left(2t^{1}+3\right)^{-\frac{1}{2}}$$

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

$$t^{0}\left(2t+3\right)^{-\frac{1}{2}}$$

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

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

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

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

Evaluate

$$\sqrt{2t+3}$$