When the function is a constant to the power of a function of x, the first step in chain rule is to rewrite f(x).  So, the first factor of f(x) will be $\displaystyle (6^{5x^2-7x})$.  Next, we have to take the derivative of the function that is the exponent, or $\displaystyle 5x^2-7x$.  Its derivative is 10x-7, so that is the next factor of our derivative.  Last, when a constant is the base of an exponential function, we must always take the natural log of that number in our derivative.  So, our final factor will be $\displaystyle ln6$.  Thus, the derivative of the entire function will be all these factors multiplied together: $\displaystyle f'(x)=6^{5x^2-7x}*(10x-7)*ln6}$.

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get $\displaystyle 12x^2-3y'=0$.  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: $\displaystyle -3y'=-12x^2$.  To get y' alone, divide both sides by -3 to get $\displaystyle y'=\frac{-12x^2}{-3}$.  To simplify even further, we can factor a -2 out of the numerator and denominator and cancel them.  So, the final answer is $\displaystyle y'=4x^2$.

To take the derivative, you must first take the derivative of the outside function, which is sine.  However, the $\displaystyle x^2$, or the angle of the function, remains the same until we take its derivative later.  The derivative of sinx is cosx, so you the first part of $\displaystyle f'(x)$ will be $\displaystyle cos(x^2)$.  Next, take the derivative of the inside function, $\displaystyle x^2$.  Its derivative is $\displaystyle 2x$, so by the chain rule, we multiply the derivatives of the inside and outside functions together to get $\displaystyle f'(x)=2xcos(x^2)$.

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get $\displaystyle 2x+2y(y')=0.$  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: $\displaystyle 2y(y')=-2x.$.  To get y' alone, divide both sides by $\displaystyle 2y$ to get $\displaystyle y'=\frac{-2x}{2y}$.  To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them.  So, the final answer is $\displaystyle \frac{-x}{y}$.

To find the derivative, we can first rewrite the function to make it easier to take the chain rule.  Rewrite $\displaystyle 10^{2\sqrt{x}}$ as $\displaystyle 10^{2x^{\frac{1}{2}}}$.  Now, like in any exponential function, the first factor of the derivative is the original exponential function.  So, the first factor of f'(x) will be $\displaystyle 10^{2\sqrt{x}}$.  Next, by the chain rule for derivatives, we must take the derivative of the exponent, which is why we rewrote the exponent in a way that is easier to take the derivative of.  So, the derivative of the exponent is $\displaystyle x^{-{\frac{1}{2}}}$, because the 1/2 and the 2 cancel when we bring the power down front, and the exponent of 1/2 minus 1 becomes negative 1/2.  The last factor of the derivative is $\displaystyle ln10$ because in every derivative of an exponential function where the base is a number, we must multiply by the natural log of that base.  So, once you multiply all these factors together, the final answer is $\displaystyle f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}$

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get $\displaystyle cos(x+y)(1+y')=y^2(-sinx)+cosx(2y*y').$  The next step is to solve for y', so we put all terms containing y' on the left side of the equation and factor out a common y': $\displaystyle y'(cos(x+y)-2ycosx)=-y^2sinx-cos(x+y)$.  To get y' alone, divide both sides by$\displaystyle cos(x+y)-2ycosx$ to get $\displaystyle y'=\frac{-y^2sinx-cos(x+y)}{cos(x+y)-2ycosx}$.

By the chain rule, we must first take the derivative of the outside function by bringing the power down front and reducing the power by one.  When we do this, we do not change the function that is in the parentheses, or the inside function.  That means that the first part of $\displaystyle f'(x)$ will be $\displaystyle 2(4x^3-9x^2+4x-8)$.  Next, we must take the derivative of the inside function.  Its derivative is $\displaystyle (12x^2-18x+4)$.  The chain rule says we must multiply the derivative of the outside function by the derivative of the inside function, so the final answer is $\displaystyle f'(x)=2(4x^3-9x^2+4x-8)*(12x^2-18x+4)$.

Before we take the derivative of the logarithmic function, we can make it easier for ourselves by simplifying the equation to $\displaystyle lny=6lnx$.  We can bring the exponent of 6 down in front of the natural log of x due to properties of logarithms.  Next, take the derivative of each term in terms of x.  Don't forget to multiply by y' each time you take the derivative of a term containing y!  When we do this, we should get $\displaystyle (\frac{1}{y})y'=6(\frac{1}{x})$ because the derivative of lnx is 1/x.  Next, solve for y' by multiplying both sides by y to get the final answer of $\displaystyle y'=\frac{6y}{x}$.

To take the derivative of any exponential function, we repeat the exponential function in the derivative.  So, the first factor of the derivative will be $\displaystyle e^{sec3\theta}$.  Next, we have to take the derivative of the exponent using chain rule.  The derivative of the trigonometric function secx is secxtanx, so in terms of this problem its derivative is $\displaystyle sec(3\theta)tan(3\theta)$.  Since the angle has a scalar of 3, we must also multiply the entire derivative by 3.  So, the answer is $\displaystyle y'=3e^{sec3\theta}*sec(3\theta)tan(3\theta)$.

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get $\displaystyle 2x(y)+(x^2)y'+(6y^2)y'=3+2y'$.  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: $\displaystyle x^2y'+6y^2(y')-2y'=3-2xy$.  Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: $\displaystyle y'(x^2+6y^2-2)=3-2xy$.  To get y' alone, divide both sides by $\displaystyle (x^2+6y^2-2)$ to get a final answer of  $\displaystyle y'=\frac{3-2xy}{x^2+6y^2-2}$.