3.6 chain rule\n5. calculate the following derivatives.\na. $\frac{d}{dx}sin(3x^{2}+5)$\nb…

3.6 chain rule\n5. calculate the following derivatives.\na. $\frac{d}{dx}sin(3x^{2}+5)$\nb. $\frac{d}{dx}(x^{5}+2x + 1)^{4}$\nc. $\frac{d}{dx}e^{3x^{2}}$\nd. $\frac{d}{dx}sin^{5}(x)$\ne. $\frac{d}{dx}(\frac{1}{sqrt{x^{2}+1}})$\nf. $\frac{d}{dx}(x^{2}cos^{5}x)$\ng. $\frac{d}{dx}(\tan^{5}(2pi x))$\nh. $\frac{d}{dx}sqrt3{1+\tan(pi x^{2})}$\ni. $\frac{d}{dx}(4cos^{7}(x)-10sec^{2}(x))$\nj. $\frac{d}{d\theta}sin(sin(cos(7\theta)))$\n6. (i) find $f(x)$ for $f(x)=(x^{2}+1)^{5}sin^{2}x$.

3.6 chain rule\n5. calculate the following derivatives.\na. $\frac{d}{dx}sin(3x^{2}+5)$\nb. $\frac{d}{dx}(x^{5}+2x + 1)^{4}$\nc. $\frac{d}{dx}e^{3x^{2}}$\nd. $\frac{d}{dx}sin^{5}(x)$\ne. $\frac{d}{dx}(\frac{1}{sqrt{x^{2}+1}})$\nf. $\frac{d}{dx}(x^{2}cos^{5}x)$\ng. $\frac{d}{dx}(\tan^{5}(2pi x))$\nh. $\frac{d}{dx}sqrt3{1+\tan(pi x^{2})}$\ni. $\frac{d}{dx}(4cos^{7}(x)-10sec^{2}(x))$\nj. $\frac{d}{d\theta}sin(sin(cos(7\theta)))$\n6. (i) find $f(x)$ for $f(x)=(x^{2}+1)^{5}sin^{2}x$.

Answer

Explanation:

Step1: Recall chain - rule

The chain - rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Also, for a product (y = u\cdot v), (y^\prime=u^\prime v + uv^\prime).

Step2: Solve part A

Let (u = 3x^{2}+5), then (y=\sin(u)). By the chain - rule, (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). (\frac{dy}{du}=\cos(u)) and (\frac{du}{dx}=6x[SSE Completed, Client Connection Error][LLM SSE On Failure]