# Thread: Check my derivatives solutions.

1. ## Check my derivatives solutions.

1. Find second derivative: $\displaystyle y=(x^2-5)*e^x$

Find second derivative:
$\displaystyle y1=2x*e^x+(x^2-5)*e^x$
$\displaystyle y2=(2x)*(e^x)+2(e^x)+(x^2-5)*(e^x)+(2x)*(e^x)$

Find Derivatives:
1. $\displaystyle f(x)=\frac{3-x^4}{x^2+5}$
$\displaystyle f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}$

2. $\displaystyle f(x)=(14-3x^2)^5$
$\displaystyle f1(x)= -6x*5u^4$
=$\displaystyle f1(x)=-6x*5(14-3x^2)^4$
=$\displaystyle f1(x)=-30x(14-3x^2)^4$

3. $\displaystyle f(x)=\sqrt{x^2-3}$
$\displaystyle f(x)=(x^2-3)^\frac{1}{2}$
=$\displaystyle f1(x)=2x*\frac{1}{2}(x^2-3)^\frac{-1}{2}$
=$\displaystyle f1(x)=1.5(x^2-3)^\frac{-1}{2}$

4. $\displaystyle g(x)=2Ln(x^5)$
$\displaystyle g1(x)=5x^4*\frac{2}{x^5}$ or (i think) $\displaystyle g1(x)=5x^4*\frac{5x^4}{2ln}$

5. $\displaystyle h(x)=5x^2*LnX$
$\displaystyle h1(x)=10x*Lnx+5x^2*\frac{1}{x}$
=$\displaystyle h1(x)=10x*Lnx+5x$

2. ????

3. Originally Posted by MathKidd

1. Find second derivative: $\displaystyle y=(x^2-5)*e^x$

Find second derivative:
$\displaystyle y1=2x*e^x+(x^2-5)*e^x$
$\displaystyle y2=(2x)*(e^x)+(e^x)+(x^2-5)*(e^x)+(2x)*(e^x)$
$\displaystyle y''=(2x)*(e^x)+2(e^x)+(x^2-5)*(e^x)+(2x)*(e^x)$

Originally Posted by MathKidd
Find Derivatives:
1. $\displaystyle f(x)=\frac{3-x^4}{x^2+5}$
$\displaystyle f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}$
Looks good but can be simplified

Originally Posted by MathKidd
2. $\displaystyle f(x)=(14-3x^2)^5$
$\displaystyle f1(x)= -6x*5u^4$
$\displaystyle =f1(x)=-6x*5(14-3x^2)^4$
$\displaystyle =f1(x)=-30x(14-3x^2)^4$
Looks good

Originally Posted by MathKidd
3. $\displaystyle f(x)=\sqrt{x^2-3}$
$\displaystyle f(x)=(x^2-3)^\frac{1}{2}$
$\displaystyle =f1(x)=2x*\frac{1}{2}(x^2-3)^\frac{-1}{2}$
$\displaystyle =f1(x)=1.5(x^2-3)^\frac{-1}{2}$
$\displaystyle f'(x)=x(x^2-3)^\frac{-1}{2}$

Originally Posted by MathKidd
4. $\displaystyle g(x)=2Ln(x^5)$
$\displaystyle g1(x)=5x^4*\frac{2}{x^5} or (i think) g1(x)=5x^4*\frac{5x^4}{2ln}$
$\displaystyle g'(x)=5x^4*\frac{2}{x^5}=\frac{10}{x}$
5. $\displaystyle h(x)=5x^2*LnX$
$\displaystyle h1(x)=10x*Lnx+5x^2*\frac{1}{x}$
$\displaystyle =h1(x)=10x*Lnx+5x$