Answer to Question #315707 in Calculus for Aliyah

Question

7. Differentiate the function x = y ^ 4 - 2y ^ 3

8. Differentiate the function t = 1/2 * t ^ 4 - 5t - 3

9. Differentiate the function y = (x ^ 2 - 2) ^ 2

10. Differentiate the function y = 1/(x + 7)

11. Find the slope of function y = 1/(x ^ 2), 12, 1/4 12. Find the slope of function y^ 2 =4x.(1,2)

13. Find the slope of function y = 1/(x + 1), (- 2, 1)

Answer 1

$1) x=y^4-2y^3\\ 1=4y^3\cdot y'-6y^2\cdot y'\\ 1=y'(4y^3-6y^2)\\ 2) y=\frac{1}{2}t^4-5t-3\\ y'=\frac{1}{2}\cdot 4 t^3-5=2t^3-5\\ 3) y=(x^2-2)^2\\ y'=2(x^2-2)\cdot 2x=4x(x^2-2)\\ 4)y=\frac{1}{x+7}=(x+7)^{-1}\\ y'= - (x+7)^{-2}= \frac{-1}{(x+7)^2}\\ 5) y=\frac{1}{x^2}=x^{-2}, (12,\frac{1}{4})\\ y'=k=\tan \alpha=f'(x_0)\\ y'=-2x^{-3},\\ y'(12)=-2(12)^{-3}=-\frac{1}{864}\\ \tan \alpha=-\frac{1}{864}\\ y'(\frac{1}{4})=-2(\frac{1}{4})^{-3}=-\frac{1}{32}\\ \tan \alpha=-\frac{1}{32}\\ 6)y^2=4x, (1,2)\\ y=\sqrt{4x}\\ y'=\frac{1}{2\sqrt{4x}}\cdot 4=\frac{1}{\sqrt{x}}\\ y'(1)=\frac{1}{\sqrt{1}}=1\\ \tan\alpha=1\\ \alpha=45^0\\ 7) y=\frac{1}{x+1}, (-2,1)\\ y'=-\frac{1}{(x+1)^2}\\ y'(-2)=-\frac{1}{(-2+1)^2}=-1\\ \tan \alpha=-1\\ \alpha=135^0\\$